3. CUMT
HEAT TRANSFER LECTURE
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
0
T T
x y
∂ ∂
+ =
∂ ∂
The solution to this equation may be obtained by analytical,
numerical, or graphical techniques.
(3-1)
3-1 Introduction
admission.edhole.com
4. CUMT
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
3-1 Introduction
admission.edhole.com
5. CUMT
HEAT TRANSFER LECTURE
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
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.
admission.edhole.com
6. CUMT
HEAT TRANSFER LECTURE
Consider a sine-wave distribution on the upper edge, the
boundary conditions are:
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
7. CUMT
HEAT TRANSFER LECTURE
Substitute:
We obtain two ordinary differential equations in terms of
this constant,
2 2
2 2
1 1T T
X x Y y
∂ ∂
− =
∂ ∂
2
2
2
0
X
X
x
λ
∂
+ =
∂
2
2
2
0
Y
Y
y
λ
∂
− =
∂
where λ2
is called the separation constant.
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
8. CUMT
HEAT TRANSFER LECTURE
We write down all possible solutions and then see which
one fits the problem under consideration.
( ) ( )
1 2
2
3 4
1 2 3 4
0:For X C C x
Y C C y
T C C x C C y
λ = = +
= +
= + +
This function cannot fit the sine-function boundary
condition, so that the solution may be excluded.2
0λ =
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
9. CUMT
HEAT TRANSFER LECTURE
( )( )
2
5 6
7 8
5 6 7 8
0:
cos sin
cos sin
x x
x x
For X C e C e
Y C y C y
T C e C e C y C y
λ λ
λ λ
λ
λ λ
λ λ
−
−
< = +
= +
= + +
This function cannot fit the sine-function boundary condition,
so that the solution may be excluded.2
0λ <
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
10. CUMT
HEAT TRANSFER LECTURE
( ) ( )
2
9 10
11 12
9 10 11 12
0: cos sin
cos sin
y y
y y
For X C x C x
C e C e
T C x C x C e C e
Y λ λ
λ λ
λ λ λ
λ λ
−
−
> = +
= +
= + +
It is possible to satisfy the sine-function boundary condition;
so we shall attempt to satisfy the other condition.
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
11. CUMT
HEAT TRANSFER LECTURE
Let
The equation becomes:
Apply the method of variable separation, let
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
12. CUMT
HEAT TRANSFER LECTURE
And the boundary conditions become:
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
13. CUMT
HEAT TRANSFER LECTURE
Applying these conditions,we have:
( ) ( )9 10 11 120 cos sinC x C x C Cλ λ= + +
( )9 11 120 y y
C C e C eλ λ−
= +
( ) ( )9 10 11 120 cos sin y y
C W C W C e C eλ λ
λ λ −
= + +
( ) ( )9 10 11 12sin cos sin H H
m
x
T C x C x C e C e
W
λ λπ
λ λ −
= + + ÷
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
14. CUMT
HEAT TRANSFER LECTURE
accordingly,
and from (c),
This requires that
11 12C C= −
9 0C =
( )10 120 sin y y
C C W e eλ λ
λ −
= −
sin 0Wλ =
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
15. CUMT
HEAT TRANSFER LECTURE
then
which requires that Cn =0 for n >1.
We get
n
W
π
λ =
1
1
sin sinhn
n
n x n y
T T C
W W
π π
θ
∞
=
= − = ∑
1
sin sin sinhm n
n
x n x n H
T C
W W W
π π π∞
=
= ∑
The final boundary condition may now be applied:
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
16. CUMT
HEAT TRANSFER LECTURE
The final solution is therefore
( )
( ) 1
sinh /
sin
sinh /
m
y W x
T T T
H W W
π π
π
= +
The temperature field for this problem is shown. Note that the heat-
flow lines are perpendicular to the isotherms.
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
17. CUMT
HEAT TRANSFER LECTURE
Another set of boundary conditions
0 at 0
0 at 0
0 at
sin atm
y
x
x W
x
T y H
W
θ
θ
θ
π
θ
= =
= =
= =
= = ÷
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
18. CUMT
HEAT TRANSFER LECTURE
Using the first three boundary conditions, we obtain the
solution in the form of Equation:
1
1
sin sinhn
n
n x n y
T T C
W W
π π∞
=
− = ∑
Applying the fourth boundary condition gives
2 1
1
sin sinhn
n
n x n H
T T C
W W
π π∞
=
− = ∑
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
19. CUMT
HEAT TRANSFER LECTURE
This series is
then
( )
( )
1
2 1 2 1
1
1 12
sin
n
n
n x
T T T T
n W
π
π
+
∞
=
− +
− = − ∑
( )
( )
( )
1
2 1
1 12 1
sinh /
n
nC T T
n H W nπ π
+
− +
= −
The final solution is expressed as
( ) ( )
( )
1
1
12 1
1 1 sinh /2
sin
sinh /
n
n
n y WT T n x
T T n W n H W
ππ
π π
+
∞
=
− +−
=
−
∑
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
20. CUMT
HEAT TRANSFER LECTURE
0 at 0
0 at 0
0 at
sin atm
y
x
x W
x
T y H
W
θ
θ
θ
π
θ
= =
= =
= =
= = ÷
Transform the boundary condition:
3-2 Mathematical Analysis of Two-Dimensional Heat Conduction
admission.edhole.com
22. CUMT
HEAT TRANSFER LECTURE
3-4 The Conduction Shape Factor
Consider a general one dimensional heat conduct-
ion problem, from Fourier’s Law:
let
then
where : S is called shape factor.admission.edhole.com
23. CUMT
HEAT TRANSFER LECTURE
Note that the inverse hyperbolic cosine can be calculated from
( )1 2
cosh ln 1x x x−
= ± −
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,
wall
A
S
L
= edge 0.54S D= corner 0.15S L=
where A = area of wall, L = wall thickness, D = length of edge
3-4 The Conduction Shape Factor
admission.edhole.com
35. CUMT
HEAT TRANSFER LECTURE
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.
admission.edhole.com
37. CUMT
HEAT TRANSFER LECTURE
2 、 Discrete equation
Taylor series expansion
2 2 3 3
1, , 2 3
, , ,
( ) ( )
...
2 6
m n m n
m n m n m n
T x T x T
T T x
x x x
+
∂ ∆ ∂ ∆ ∂
= + ∆ + + +
∂ ∂ ∂
2 2 3 3
1, , 2 3
, , ,
( ) ( )
...
2 6
m n m n
m n m n m n
T x T x T
T T x
x x x
−
∂ ∆ ∂ ∆ ∂
= − ∆ + − +
∂ ∂ ∂
3-5 Numerical Method of Analysis
admission.edhole.com
38. CUMT
HEAT TRANSFER LECTURE
2
2
1, 1, , 2
,
2 ( ) ...m n m n m n
m n
T
T T T x
x
+ −
∂
+ = + ∆ +
∂
2
1, , 1, 2
2 2
,
2
( )
( )
m n m n m n
m n
T T TT
o x
x x
+ −− +∂
= + ∆
∂ ∆
2
2
1,,1,
,
2
2
)(
)(
2
yo
y
TTT
y
T nmnmnm
nm
∆+
∆
+−
=
∂
∂ −+
2 、 Discrete equation
Differential equation for two-dimensional steady-state heat flow
2 2
2 2
0
T T q
x y k
•
∂ ∂
+ + =
∂ ∂
3-5 Numerical Method of Analysis
admission.edhole.com
39. CUMT
HEAT TRANSFER LECTURE
2 、 Discrete equation
Discrete equation at nodal point (m,n)
1, , 1, , 1 , , 1
2 2
2 2
0m n m n m n m n m n m nT T T T T T q
x y k
•
+ − + −− + − +
+ + =
∆ ∆
1, , 1, , 1 , , 1
2 2
2 2
0m n m n m n m n m n m nT T T T T T
x y
+ − + −− + − +
+ =
∆ ∆
no heat generation
Δx= Δy
1, 1, , 1 , 1 ,4 0m n m n m n m n m nT T T T T+ − + −+ + + − =
3-5 Numerical Method of Analysis
admission.edhole.com
40. CUMT
HEAT TRANSFER LECTURE
2 、 Discrete equation
Thermal balance
(1) Interior points
steady-state & no heat
generation
1, 1, , 1 , 1 0m n m n m n m nq q q q− + + −+ + + =
1, ,
1,
d
d
m n m n
m n
T TT
q kA k y
x x
−
−
−
= − = ∆
∆
x
TT
ykq nmnm
nm
∆
−
∆= +
+
,,1
,1
, 1 ,
, 1
m n m n
m n
T T
q k x
y
+
+
−
= ∆
∆
, 1 ,
, 1
m n m n
m n
T T
q k x
y
−
−
−
= ∆
∆
1, , 1, , , 1 , , 1 ,
0m n m n m n m n m n m n m n m nT T T T T T T T
k y k y k x k x
x x y y
− + + −− − − −
∆ + ∆ + ∆ + ∆ =
∆ ∆ ∆ ∆
3-5 Numerical Method of Analysis
admission.edhole.com
41. CUMT
HEAT TRANSFER LECTURE
Thermal balance
Δx= Δy
1, 1, , 1 , 1 ,4 0m n m n m n m n m nT T T T T+ − + −+ + + − =
genq q V q x y
• •
= × = ×∆ ∆
2
1, 1, , 1 , 1 ,4 ( ) 0m n m n m n m n m n
q
T T T T T x
k
•
+ − + −+ + + − + ∆ =
steady-state with heat generation
(1) Interior points
3-5 Numerical Method of Analysis
admission.edhole.com
42. CUMT
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
−
−
−
= ∆
∆
, 1 ,
, 1
2
m n m n
m n
T Tx
q k
y
+
+
−∆
=
∆
, 1 ,
, 1
2
m n m n
m n
T Tx
q k
y
−
−
−∆
=
∆
,( )w m nq h y T T∞= ×∆ × −
1, , , 1 , , 1 ,
,( )
2 2
m n m n m n m n m n m n
m n
T T T T T Tx x
k y k k h y T T
x y y
− + −
∞
− − −∆ ∆
∆ + + = ×∆ × −
∆ ∆ ∆
3-5 Numerical Method of Analysis
admission.edhole.com
43. CUMT
HEAT TRANSFER LECTURE
Thermal balance
Δx= Δy
(2) boundary
points
, 1, , 1 , 1
1
( 2) ( )
2
m n m n m n m n
h x h x
T T T T T
k k
− + − ∞
×∆ ×∆
+ = + + +
1, , , 1 ,
, ,( ) ( )
2 2 2 2
m n m n m n m n
m n m n
T T T Ty x x y
k k h T T h T T
x y
− −
∞ ∞
− −∆ ∆ ∆ ∆
+ = × × − + × × −
∆ ∆
, 1, , 1
1
( 1) ( )
2
m n m n m n
h x h x
T T T T
k k
− − ∞
×∆ ×∆
+ = + +
Δx= Δy
3-5 Numerical Method of Analysis
admission.edhole.com
44. CUMT
HEAT TRANSFER LECTURE
Thermal balance
(2) boundary
points
Δx= Δy
1, , 1, , , 1 ,
, 1 ,
, ,
2 2
( ) ( )
2 2
m n m n m n m n m n m n
m n m n
m n m n
T T T T T Ty x
k y k k
x x y
T T x y
k x h T T h T T
y
− + −
+
∞ ∞
− − −∆ ∆
∆ + +
∆ ∆ ∆
− ∆ ∆
+ ∆ = × × − + × × −
∆
, 1, 1, , 1 , 1
1
( 3) (2 2 )
2
m n m n m n m n m n
h x h x
T T T T T T
k k
− + − + ∞
×∆ ×∆
+ = + + + +
3-5 Numerical Method of Analysis
admission.edhole.com
45. CUMT
HEAT TRANSFER LECTURE
3 、 Algebraic equation
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
......
......
............................................
......
n n
n n
n n nn n n
a T a T a T C
a T a T a T C
a T a T a T C
+ + + =
+ + + =
+ + + =
3-5 Numerical Method of Analysis
admission.edhole.com
46. CUMT
HEAT TRANSFER LECTURE
Matrix notation
[ ]
11 12 1
21 22 2
1 2
...
...
... ... ... ...
...
n
n
n n nn
a a a
a a a
A
a a a
=
[ ]
1
2
...
n
T
T
T
T
=
[ ]
1
2
...
n
C
C
C
C
=
[ ][ ] [ ]A T C=
Iteration
Simple Iteration & Gauss-Seidel Iteration
3-5 Numerical Method of Analysis
admission.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.
admission.edhole.com
48. CUMT
HEAT TRANSFER LECTURE
[Solution]
The equations for nodes 1,2,4,5 are given by
2 4 1
1 3 5 2
1 5 7 4
2 4 6 8 5
500 100 4 0
500 4 0
100 4 0
4 0
T T T
T T T T
T T T T
T T T T T
+ + + − =
+ + + − =
+ + + − =
+ + + − =
Example 3-5
admission.edhole.com
49. CUMT
HEAT TRANSFER LECTURE
[Solution]
Equations for nodes 3,6,7,8 are
The equation for node 9 is
9 6 8
1 1 1
1 ( ) 100
3 2 3
T T T= + + ×
3 2 6
6 3 5 9
7 4 8
8 7 5 9
1 1 1
2 (500 2 ) 100
3 2 3
1 1 1
2 ( 2 ) 100
3 2 3
1 1 1
2 (100 2 ) 100
3 2 3
1 1 1
2 ( 2 ) 100
3 2 3
T T T
T T T T
T T T
T T T T
= + + + ×
= + + + ×
= + + + ×
= + + + ×
Example 3-5
admission.edhole.com
50. CUMT
HEAT TRANSFER LECTURE
3 2 6
1 1 1
2 (500 2 ) 100
3 2 3
T T T= + + + ×
6 3 5 9
1 1 1
2 ( 2 ) 100
3 2 3
T T T T= + + + ×
7 4 8
1 1 1
2 (100 2 ) 100
3 2 3
T T T= + + + ×
8 7 5 9
1 1 1
2 ( 2 ) 100
3 2 3
T T T T= + + + ×
The equation for node 9 is
9 6 8
1 1 1
1 ( ) 100
3 2 3
T T T= + + ×
Example 3-5
admission.edhole.com
51. CUMT
HEAT TRANSFER LECTURE
We thus have nine equations and nine unknown nodal temperatures. So
the answer is
For the 500 face, the heat flow into the face is℃
31 2 (500 )(500 ) (500 )
10 [ ]
2
... 4843.4 /
in
TT TT x
q k A x x
y y y y
W m
−− −∆ ∆
= × × = × ∆ × + ∆ × + ×
∆ ∆ ∆ ∆
= =
∑
The heat flow out of the 100 face is℃
71 4
1
( 100)( 100) ( 100)
10 [ ]
2
... 3019 /
TT TT y
q k A y y
x x x x
W m
−− −∆ ∆
= × × = × ∆ × + ∆ × + ×
∆ ∆ ∆ ∆
= =
∑
Example 3-5
admission.edhole.com
52. CUMT
HEAT TRANSFER LECTURE
2
3 6 9
( )
10 [ ( 100) ( 100) ( 100)]
2
... 1214.6 /
q h A T T
y
y T y T T
W m
∞= × × −
∆
= × ∆ × − + ∆ × − + × −
= =
∑
The heat flow out the right face is
3
7 8 9
( )
10 [ ( 100) ( 100) ( 100)]
2
... 600.7 /
q h A T T
y
y T y T T
W m
∞= × × −
∆
= × ∆ × − + ∆ × − + × −
= =
∑
1 2 3 ... 4834.3 /outq q q q W m= + + = = 4843.4 /inq W m=<
The heat flow out the bottom face is
The total heat flow out is
Example 3-5
admission.edhole.com
53. CUMT
HEAT TRANSFER LECTURE
3-6 Numerical Formulation in Terms of Resistance Elements
Thermal balance — the net heat input to node i must be zero
0j i
i
j i j
T T
q
R
−
+ =∑
qi — heat generation, radiation, etc.
i — solving node
j — adjoining node
admission.edhole.com
54. CUMT
HEAT TRANSFER LECTURE
1, , 1, , , 1 , , 1 ,
0m n m n m n m n m n m n m n m n
m m n n
T T T T T T T T
R R R R
− + − +
− + − +
− − − −
+ + + =
x
R
kA
∆
=
1
m m n nR R R R
k
− + − += = = =
1, 1, , 1 , 1 ,4 0m n m n m n m n m nT T T T T+ − + −+ + + − =
0
j i
i
j i j
T T
q
R
−
+ =∑
so
3-6 Numerical Formulation in Terms of Resistance Elements
admission.edhole.com
55. CUMT
HEAT TRANSFER LECTURE
3-7 Gauss-Seidel Iteration
0
j i
i
j i j
T T
q
R
−
+ =∑
( / )
(1/ )
i j i j
j
i
i j
j
q T R
T
R
+
=
∑
∑
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.
admission.edhole.com
56. CUMT
HEAT TRANSFER LECTURE
Convergence Criterion
( 1) ( )i n i nT T δ+ − ≤
( 1) ( )
( )
i n i n
i n
T T
T
ε+ −
≤
3 6
10 10ε − −
= ~
Biot number
h x
Bi
k
∆
=
3-7 Gauss-Seidel Iteration
admission.edhole.com
57. CUMT
HEAT TRANSFER LECTURE
Example 3-6
1x
R
k y k
∆
= =
∆
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
Therefore, we have
( / ) ( ) ( )
(1/ ) ( ) ( )
i j i j i j j j j
j j j
i
i j j j
j j j
q T R q k T k T
T
R k k
+ +
= = =
∑ ∑ ∑
∑ ∑ ∑admission.edhole.com
58. CUMT
HEAT TRANSFER LECTURE
Example 3-6
Because each node has four resistance connected to it and k is assumed
constant, so
4j
j
k k=∑
1
4
i j
j
T T∴ = ∑
admission.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△
admission.edhole.com
60. CUMT
HEAT TRANSFER LECTURE
Summary
(1)Numerical Method
Solving Zone
Nodal equations
thermal balance method — Interior & boundary point
Algebraic equations
Gauss-Seidel iteration
( / )
(1/ )
i j i j
j
i
i j
j
q T R
T
R
+
=
∑
∑
admission.edhole.com
61. CUMT
HEAT TRANSFER LECTURE
Summary
(2)Resistance Forms
0j i
i
j i j
T T
q
R
−
+ =∑
(3)Convergence
Convergence Criterion
( 1) ( )i n i nT T δ+ − ≤
( 1) ( )i n i nT T δ+ − ≤
admission.edhole.com