Download - Chapter 7 Heat Conduction
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!"#$%&' )
*%+,-*%#%& .% !/0+12/0
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6-1 What is Heat?
Heat is a form of energy that exist by virtue of temperature difference.
Heat is transferred (flows) from a higher-temperature to a lower-
temperature region.
Note: Heat transfer can take place by: a) conduction, b) convection andc) radiation, in a steady-state and in a transient conditions.
Steady-state conduction occurs when the temperature at all points in thesolid body does not vary with time.
Three modes of heat transfer.
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Another illustration of modes of heat transfer.
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Actual scenario of heat conduction through a glass window.
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6-2 One-Dimensional Steady-State Conduction
We will focus on the one-dimensional steady-state conduction problems
only. It is the easiest heat conduction problem.
In one-dimensional problems, temperature gradient exists along one
coordinate axis only.
Objective
The objective of our analysis is to determine; a) the temperature distributionwithin the body and, b) the amount of heat transferred (heat flux).
1T
2T
3T
xq
x
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An energy balance across a control volume (shaded area) yields,
AdxdxdqqQAdxqA !
"#$
%& +=+
6-3 The Governing Equation
Consider heat conduction q (W/m2) through a plane wall, in which there is a
uniform internal heat generation, Q (W/m3).
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where q = heat flux per unit area (W/m2)
A = area normal to the direction of heat flow (m2)Q = internal heat generated per unit volume (W/m3)
Cancelling term qA and rearranging, we obtain,
dx
dqQ =
For one-dimensional heat conduction, the heat flux q is governed by the
Fourier’s law, which states that,dT
q k dx
! "= # $ % &
' (
where k = thermal conductivity of the material (W/m.K)
(dT/dx) = temperature gradient in x-direction (K/m)
Note: The –ve sign is due to the fact that heat flows from a high-temperature tolow- temperature region.
… (i)
… (ii)
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Fourier established the partial differential
equation governing heat diffusion process.
He solved the equation using infinite series
of trigonometric functions.
He introduced the representation of afunction as a series of sine or cosines, now
known as the Fourier series.
Jean Baptiste Joseph Fourier March 21 1768 - May 16 1830
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Substituting eq.(ii) into eq.(i) yields,
0=+!"
#$%
&Q
dx
dT k
dx
d
The governing equation has to be solved with appropriate boundary conditions
to get the desired temperature distribution, T .
Note:
Q is called a source when it is +ve (heat is generated), and is called a sink whenit is -ve (heat is consumed).
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6-4 Boundary Conditions
There are three types of thermal boundary conditions:
a) Specified temperature, T i = T o;
b) Specified heat flux, e.g., qi = 0 (insulated edge or surface);
c) Convection at the edge or surface, (h & T ! are specified).
These are illustrated below.
Note: h is the convective heat transfer coefficient (W/m2K).
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6-5 Finite Element Modeling
The uniform wall can be modeled using one-
dimensional element.
To obtain reasonably good temperature
distribution, we will discretize the wall into
several 1-D heat transfer elements, as shown.
Note:
X represents the global coordinate system.
Can you identify the kind of boundary
conditions present?
There is only one unknown quantity at any
given node, i.e. the nodal temperature, T i.
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For a one-dimensional steady-state conduction, temperature varies linearly along
the element.
Therefore we choose a linear temperature function given by,( )
2211 T N T N T +=! or ( ) [ ]T N T =!
6-6 Temperature Function
For a given element in local coordinate (! ), temperature T varies along the
length of the element.
We need to establish a temperature function so that we can obtain the
temperature T , at any location along the element, by interpolation.
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( ) ( )
( )12 1 2 1
2 21 d x x x x dx x x
! ! = " " # =" "
We wish to express the (dT/dx ) term in the governing equation in terms of
element length, l e, and the nodal temperature vector, {T }. Using the chain
rule of differentiation
!
! !
! d
dT
dx
d
dx
d
d
dT
dx
dT "="=
( ) ( ) ( ) 2121
2
1
2
11
2
11
2
1T T
d
dT T T T +!="++!=
# # # #
Substitute eq.(ii) and eq.(iii) into eq.(i) we get,
( )2112
21
12
1
2
1
2
12T T
x xT T
x xdx
dT +!
!="
#$%
&' +!
!=
where ( )! "= 12
1
1 N ( )! += 1
2
1
2 N and
Recall, …(ii)
…(iii)
…(i)
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or, [ ]{ }
[ ]{ }
2 1
11 1
e
e
T
dT T
dx x x
dT B T
dx
=
!!
=
where [ ]
( )
[ ] [ ]2 1
1 11 1 1 1
T
e
B
x x l
= ! = !
!
is called the temperature-gradient matrix. The heat flux, q (W/m2) can then
be expressed as
[ ] 12
11 1
e
T q k
T l
! "= # $ # % &
' (
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The element conductivity matrix [k T ] for the 1-D heat transfer element
will be derived using the potential energy approach.
Recall, the conduction governing equation with internal heat generation,
0=+!"
#$%
&Q
dx
dT k
dx
d
Imposing the following two boundary conditions,
( )!==
"== T T hqT T L L xo x and0
6-7 Element Conductivity Matrix
and solving the equation yields the total potential energy, !T given by
( )2
2
0 0
1 1
2 2
L L
T L
dT k dx QTdx h T T
dx! "
# $= % + %
& '( ) * *
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2 1
2 1
2
2 2
el x xd dx dx d d x x
! ! ! "= # = ="
[ ]{ } [ ]{ }( ) ( )
ande e
T
dT T N T B T
dx= =
Assuming that heat source Q = Qe and thermal conductivity k = k e are constant
within the element, the functional " T becomes
{ } [ ] [ ] { }
[ ] { } ( )
1( ) ( )
1
1 ( ) 2
1
1
2 2
1
2 2
e T ee eT T T
e
ee e L
e
k l T B B d T
Q l N d T h T T
! "
"
#
$#
% &= #' (
) *
% &+ #' (
) *
+ ,
+ ,
Note: The first term of the above equation is equivalent to the internal strain
energy for structural problem. We identify the element conductivity matrix,
[ ] [ ] [ ]1
12
T e e
T T T
k l k B B d !
"= # $
Substitute for dx and (dT/dx) in terms of ! and {T }e,
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Solving the integral and simplifying yields the element conductivity
matrix, given by
[ ] 1 1
1 1
e
T
e
k k
l
!" #= $ %!& '
Note: If the finite element model comprises of more than one element, then the
global conductivity matrix can be assembled in usual manner to give
[ ]
11 12 1
21 22 2
1 2 ...
L
L
T
L L LL
K K K
K K K K
K K K
! "# $# $
=
# $# $% &
!
!
"
(W/m2K)
(W/m2K)
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Exercise 6-1
A composite wall is made of material A and B as shown. Inner surface ofthe wall is insulated while its outer surface is cooled by water stream with
T !
= 30°C and heat transfer coefficient, h = 1000 W/m2K. A uniform heat
generation, Q A = 1.5 x 106 W/m3 occurs in material A. Model the wall
using two 1-D heat transfer elements.
Question: Assemble the global conductivity matrix, [ K T ].
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If there is an internal heat generation, Qe (W/m3) within the element,then it can be shown that the element heat rate vector due to the
internal heat generation is given by
{ } 2
1 W
12 m
ee e
Q
Q l r
! "#= $ %
& 'Note:
1. If there is no internal heat generation in the element, then the heat rate vector
for that element will be,
2. If there are more than one element in the finite element model, the global heatrate vector , { RQ} is assembled in the usual manner.
6-8 Element Heat Rate Vector
{ } ( )2
1 00 W
1 02 m
e e
Q
l r
! " # " #= =$ % $ %
& ' & '
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111 12 1 1
221 22 2 2
1 2 ...
Q L
Q L
QL L L LL L
R K K K T
R K K K T
R K K K T
! "# $ ! "% %& ' % %
% % % %& ' =( ) ( )
& ' % % % %
& ' % % % %* + , - , -
!
!
"" "
6-9 Global System of Linear Equations
The generic global system of linear equation for a one-dimensional steady-state heat conduction can be written in a matrix form as
Note:
1.
At this point, the global system of linear equations have no solution.
2.
Certain thermal boundary condition need to be imposed to solve the equations
for the unknown nodal temperatures.
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Exercise 6-2
Reconsider the composite wall in Exercise 6-1. a) Assemble the globalheat rate vector, { RQ}; b) Write the global system of linear equations for
the problem.
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111 12 1 11
221 22 2 2 21
1 2 1...
Q L
Q L
QL L L LL L L
R K K K K
R K K K T K
R K K K T K
! !
!
!
" #$ % " # " #& &' ( & & & &
& & & & & &' ( = )* + * + * +
' ( & & & & & &' ( & & & & & &, - . / . /. /
!
!
"" " "
Suppose uniform temperature T = # °C is specified
at the left side of a plane wall.
To impose this boundary condition, modify the
global SLEs as follows:
1.
Delete the 1st row and 1st column of [ K T ] matrix;
2.
Modify the { RQ} vector as illustrated.
Note: Make sure that you use a consistent unit.
6-10 Temperature Boundary Condition
x
L
1
oT C ! =
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( ) ( )
111 12 1 1
221 22 2 2
1 2 ...
Q L
Q L
L L LL QL L
R K K K T
R K K K T
K K K h R hT T !
" #$ % " #& &' ( & &
& & & &' (=) * ) *' ( & & & &' (
& & & &+ ++ ,- . + ,
!
!
"" "
Suppose that convection occurs on the right side of a
plane wall, i.e. at x = L.
The effect of convection can be incorporated bymodifying the global SLEs as follows:
1.
Add h to the last element of the [ K T ] matrix;
2.
Add (hT !
) to the last element of { RQ} vector.
Note: Make sure that you use a consistent unit.
6-11 Convection Boundary Condition
x
L
We get,
;T h!
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Once the temperature distribution within the wall is known, the heat flux
through the wall can easily be determined using the Fourier’s law.
We have,
Note:
1. At steady-state condition, the heat flux through all elements has the same
magnitude.
2. T 1 and T 2 are the nodal temperatures for an element.
3. l e is the element length.
6-12 The Heat Flux
[ ] 12
11 1
e
T q k
T l
! "= # $ # % &
' (W/m2
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Exercise 6-3
Reconsider the composite wall problem in Exercise 6-2. a) Impose theconvection boundary conditions; b) Solve the reduced SLEs, determine
the nodal temperatures; c) Estimate the heat flux, q through the
composite wall.
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Exercise 6-3: Nastran Solution
413 K
407 K
388 K
378 K
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( )111 12 1 1221 22 2 2
1 2
0
... 0
Q L o
Q L
QL L L LL L
R K K K T q
R K K K T
R K K K T
! " ! "#$ % ! "& & & &
' ( & && & & & & &' ( = +) * ) * ) *' ( & & & & & &' ( & & & & & &+ , - . - .- .
!
!
"" " " "
Suppose heat flux q = qo W/m2 is specified at the leftside of a plane wall, i.e. at x = 0.
The effect of specified heat flux is incorporated into the
analysis by modifying the global SLEs, as shown.
6-13 Heat Flux Boundary Condition
x
L
0q q=
Note:
q0 is input as +ve value if heat flows out of the body and as –ve value if heat is
flowing into the body. Do not alter the negative sign in the global SLEs above.
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Exercise 6-4
Reconsider the composite wall problem in Exercise 6-3. Suppose there isno internal heat generation in material A. Instead, a heat flux of q = 1500
W/m2 occurs at the left side of the wall.
Write the global system of linear equations for the plane wall and impose
the specified heat flux boundary condition.
75 W/m K A
k = !
21500 W/mq =
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Exercise 6-4: Nastran Solution
357 K
347 K
337 K
333 K
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Example 6-1
A composite wall consists of three
layers of materials, as shown. The
ambient temperature is T o = 20oC.
Convection heat transfer takesplace on the left surface of the wall
where T ! = 800
o
C and h = 25 W/m2oC.
Model the composite wall using
three heat transfer elements and
determine the temperature
distribution in the wall.
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Example 6-1: Nastran Solution
305.8 C
120.5 C
54.6 C
20 C
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Solution
1. Write the element conductivity matrices
[ ]( )
[ ]( )
[ ]( )
1 3
2 2
2
2
1 1 1 120 W 50 W ;
1 1 1 10.3 m 0.15 m
1 130 W
1 10.15 m
T T o o
T o
k k C C
k C
! !" # " #= =$ % $ %! !& ' & '
!" #= $ %!& '
2. Assemble the global conductivity matrix
[ ]2
1 1 0 0
1 4 3 0 W66.7
0 3 8 5 m
0 0 5 5
T o K
C
!" #$ %! !$ %
=
$ %! !$ %
!& '
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3. Write the global system of linear equations
[ ]{ } { }T Q K T R=
!!
"
!!
#
$
!!
%
!!
&
'
=
!!
"
!!
#
$
!!
%
!!
&
'
((((
)
*
++++
,
-
.
..
..
.
4
3
2
1
4
3
2
1
5500
5830
0341
0011
7.66
R
R
R
R
T
T
T
T
4. Write the element heat rate vector
Since there is NO internal heat generation, Q in the wall, the heat rate vector
for all elements are
{ } { } { }1 2 3 0
0Q Q Q
r r r ! "
= = = # $
% &
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5. Write the global system of linear equations
1
2
3
4
1 1 0 0 0
1 4 3 0 066.7
0 3 8 5 0
0 0 5 5 0
T
T
T
T
! " #$ % " #& &' ( & &! ! & & & &' (
=) * ) *' (! ! & & & &' ( & & & &!+ , - .- .
6. Impose convection and specified temperature boundary conditions (T 4 = 20°C)results in modified system of linear equations
1
2
3
4
1.375 1 0 0 (25 800)
1 4 3 0 066.7
0 3 8 5 0 ( 5 66.7) 20
0 0 5 5 0
T
T
T
T
! "# $% & # $' '( ) ' '! ! ' ' ' '( )
=* + * +( )! ! ! ! " "' ' ' '( ) ' ' ' '!, - . /. /
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7. Solving the modified system of linear equations yields
1
2
3
4
304.6
119.0
57.1
20.0
o
T
T C
T
T
! " ! "# # # ## # # #
=$ % $ %# # # ## # # #& '& '
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Example 6-2
Heat is generated in a large plate (k = 0.8 W/moC) at a rate of 4000 W/m3.The plate is 25 cm thick. The outside surfaces of the plate are exposed to
ambient air at 30oC with a convection heat transfer coefficient of 20 W/m2oC.
Model the wall using four heat transfer elements and determine: (a) the
temperature distribution in the wall, (b) heat flux, and (c) heat loss from the
right side of the wall surface.
o
o
o
W0.8
m C
W20
m C
30 C
k
h
T !
=
=
=
Data:
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Example 6-2: Nastran Solution
84.3 C
94 C
84.3 C
55 C55 C
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Solution
[ ]( )
[ ]
( )
1
2
2
2
12.8 12.8 W
12.8 12.8 m
12.8 12.8 W
12.8 12.8 m
T o
T o
k C
k
C
!" #= $ %!& '
!" #=
$ %!& '
1. Element conductivity matrices.
Since the element length and thermal conductivity are the same for all
elements,
we have [ ]( )
[ ]
( )
3
2
4
2
12.8 12.8 W
12.8 12.8 m
12.8 12.8 W
12.8 12.8 m
T o
T o
k C
k
C
!" #= $ %!& '
!" #=
$ %!& '
1 2 3 4 5
T1 T2 T3 T4 T5 h, T $
h, T $
x
The finite element model for the plane wall is shown below.
1 2 3 4
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[ ]
12.8 12.8 0 0 0
12.8 25.6 12.8 0 0
0 12.8 25.6 12.8 0
0 0 12.8 25.6 12.8
0 0 0 12.8 12.8
T K
!" #$ %! !$ %$ %= ! !$ %
! !$ %
$ %!& '
2. Assemble the global conductivity matrix,
1 2 3 4 5
Note: Connectivity with the global node numbers is shown.
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3. Heat rate vector for each element
Since the magnitude of internal heat generation and length of all
elements are the same, we have
{ }( )
{ }
( )
{ }( )
{ }( )
1
2
3
4
1 1254000 0.0625
1 1252
1 1254000 0.0625
1 1252
1 1254000 0.0625
1 1252
1 1254000 0.0625
1 1252
Q
Q
Q
Q
r
r
r
r
! " ! "#= =$ % $ %
& ' & '
! " ! "#= =$ % $ %
& ' & '
! " ! "#= =$ % $ %
& ' & '
! " ! "#= =$ % $ %
& ' & '
{ } 2
125
250W
250m
250
125
Q R
! "# ## ## #
= $ %# ## ## #& '
4. Assemble the global heat rate
vector, we get
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5. Write the system of linear equation, [ ]{ } { }T Q K T R=
!!!
"
!!!
#
$
!!!
%
!!!
&
'
=
!!!
"
!!!
#
$
!!!
%
!!!
&
'
((((((
)
*
++++++
,
-
.
..
..
..
.
125
250
250
250
125
8.128.12000
8.126.258.1200
08.126.258.120
008.126.258.12
0008.128.12
5
4
3
2
1
T
T
T
T
T
6. Impose convection boundary conditions on both sides of the wall,
( )
( )!!!
"
!!!
#
$
!
!!
%
!!!
&
'
+
+
=
!
!!
"
!!!
#
$
!
!!
%
!!!
&
'
(
(((((
)
*
+
+++++
,
-
+.
..
..
..
.+
3020125
250
250
250
3020125
208.128.12000
8.126.258.1200
08.126.258.120
008.126.258.12
0008.12208.12
5
4
3
2
1
T
T
T
T
T
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7.
Solving the modified system of linear equations by using Gaussian
elimination method, we obtain the temperatures at the global nodesas follows,
1
2
o
3
4
5
55.0
84.3
C94.0
84.3
55.0
T
T
T
T
T
! " ! "# # # ## # # ## # # #
=$ % $ %# # # ## # # ## # # #& '& '
1 2 3 4 5
T1 T2 T3 T4 T5 h, T $
h, T $
x
Note: Notice the symmetry of the temperature distribution.
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8. Compute the heat flux and heat loss.
a) Heat flux through the wall
Consider the 4th element. Using the Fourier’s law, we have
[ ]
[ ]
1
2
2
11 1
84.310.8 1 1
55.00.0625
375m
e
T q k
T l
q
W q
! "= # $ # % &
' (
! "= # $ # % &
' (
=
b) Heat loss from the right side of the wall, per unit surface area.
Using the Newton’s law of cooling , we have
( ) ( ) 220 55 30 500 mwall W
q h T T != " = # " =
The heat flux through the
wall is not constant due to
the heat generation Q that
occurs in the wall.