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Chapter 3 Steady-State Conduction Multiple Dimensions
CUMT
HEAT TRANSFER LECTURE
CHAPER 3
Steady-State Conduction
Multiple Dimensions
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CUMT
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.
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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
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3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
CUMT
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.
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CUMT
Consider a sine-wave distribution on the upper edge, the
boundary conditions are:
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CUMT
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.
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CUMT
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.
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l l
< = +
Y C y C y
T C e C e C y C y
CUMT
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.
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= +
= + +
y y
l > = l +
l
Y l l
-
l l
-
T C x C x C e C e
CUMT
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.
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CUMT
Let
The equation becomes:
Apply the method of variable separation, let
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CUMT
And the boundary conditions become:
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CUMT
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
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CUMT
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
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CUMT
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
=
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CUMT
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.
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CUMT
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
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CUMT
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
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CUMT
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
=
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CUMT
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
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CUMT
3-3 Graphical Analysis
neglect
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CUMT
3-4 The Conduction Shape Factor
Consider a general one dimensional heat conduct-ion
problem, from Fourier’s Law:
let
then
Videow.ehdheorlee.c：om S is called shape factor.

CUMT
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
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CUMT
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CUMT
Example 3-1
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CUMT
Example 3-2
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CUMT
Example 3-3
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CUMT
Example 3-4
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CUMT
3-5 Numerical Method of Analysis
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.
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CUMT
1、Discretization of the solving
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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
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
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CUMT
2
+ = + 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
¶ + ¶ + =
¶ ¶
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CUMT
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 + - + - + + + - =
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D + D + D + D =
CUMT
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

CUMT
Thermal balance
(1) Interior points
Δx= Δy
· ·
= ´ = ´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
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CUMT
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
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x y y
¥
- D - D - D + + = ´D ´ -
D D D

CUMT
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
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CUMT
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
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CUMT
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
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é ù
ê ú
= ê ú
ê ú
ê ú
ë û
CUMT
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
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CUMT
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.
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CUMT
Example 3-5
[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
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CUMT
Example 3-5
[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
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CUMT
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
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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
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CUMT
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
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3-6 Numerical Formulation in Terms of Resistance Elements
CUMT
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
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3-6 Numerical Formulation in Terms of Resistance Elements
CUMT
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
m m n n R R R R
- + - + k = = = =
so
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CUMT
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.
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3-7 Gauss-Seidel Iteration
+ e -
£ e =10-3～10-6
CUMT
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
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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
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

CUMT
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
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CUMT
3-8 Accuracy Consideration
Truncation Error — Influenced by difference scheme
Discrete Error — Influenced by truncation error & △x
Round-off Error — Influenced by △x
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CUMT
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
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CUMT
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 + - £
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CUMT
Summary
(4)Accuracy
Truncation Error
Discrete Error
Round-off Error
Important conceptions
Nodal equations — thermal balance method
Calculated temperature & heat flow
Convergence criterion
How to improve accuracy
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CUMT
Exercises
Exercises: 3-16, 3-24, 3-48, 3-59
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