MBA HEAT TRANSFER LECTURE

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CUMT
HEAT TRANSFER LECTURE
Chapter 3 Steady-State Conduction Multiple Dimensions
CHAPER 3
Steady-State Conduction
Multiple Dimensions
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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
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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
3-1 Introduction
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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.
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Consider a sine-wave distribution on the upper edge, the
boundary conditions are:
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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.
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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 =
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( )( )
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 <
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( ) ( )
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.
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Let
The equation becomes:
Apply the method of variable separation, let
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And the boundary conditions become:
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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 − 
= + + ÷
 
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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 =
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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:
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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.
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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
= =
= =
= =
 
= = ÷
 
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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∞
=
− = ∑
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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
+
∞
=
− +−
=
−
∑
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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:
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3-3 Graphical Analysis
neglect
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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.Video.edhole.com

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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
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Example 3-1
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Example 3-2
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Example 3-3
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Example 3-4
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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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1 、 Discretization of the solving
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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
−
∂ ∆ ∂ ∆ ∂
= − ∆ + − +
∂ ∂ ∂
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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
∆+
∆
+−
=
∂
∂ −+
Differential equation for two-dimensional steady-state heat flow
2 2
2 2
0
T T q
x y k
•
∂ ∂
+ + =
∂ ∂
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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+ − + −+ + + − =
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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
− + + −− − − −
∆ + ∆ + ∆ + ∆ =
∆ ∆ ∆ ∆
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Thermal balance
Δx= Δy
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
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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
− + −
∞
− − −∆ ∆
∆ + + = ×∆ × −
∆ ∆ ∆
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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
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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
− + − + ∞
×∆ ×∆
+ = + + + +
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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
+ + + =
+ + + =
+ + + =
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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
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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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[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
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[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
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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
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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
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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
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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
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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
− + − += = = =
0
j i
i
j i j
T T
q
R
−
+ =∑
so
3-6 Numerical Formulation in Terms of Resistance Elements
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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.
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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
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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
+ +
= = =
∑ ∑ ∑
∑ ∑ ∑Video.edhole.com

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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∴ = ∑
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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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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
+
=
∑
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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+ − ≤
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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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Exercises
Exercises: 3-16, 3-24, 3-48, 3-59
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