chapters 2 heat conduction pp
TRANSCRIPT
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Chapters 2. Heat Conduction Equation
2.1 Introduction
Heat flow across various layers in a chip is governed by heatconduction equation. To understand heat conductionequation is a necessary step for calculating the temperaturedrop across each layers. In this chapter, the mean objectivesare:
To derive the heat conduction equation
To study the associated boundary conditions
To study the solutions of the steady state, one-dimensional
heat conduction problems
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2-2 The general heat conduction equation
(1) Heat conduction across an elemental volume, T = f(x,y,z,t)
is the rate of heat flow into the elemental control volume
dxdydzacross the element surface dydzatx.
The rate out flow of heat across the element surface dydzat
x+dxis
xQ&
x
Q& x dxQ
&
y
z
xdx
dydz
[( ) ( ) ]x x dx x x xQ T T
Q Q dQ Q dx k k dx dydz x x x x
x x x x! ! !
x x x x
&& & & &
( )xT
Q k dydz x
x!
x
&
x x+dx
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(2) Internal heat generation in the elemental volume
The rate of heat generation inside in volume is
is the rate of energy generation per unit volume (W/m3)
(3) Applying the principle of conservation of energy for a closed system,
the net heat input to the elemental volume is equal to the increase of
internal energy.
The general heat conduction equation is
( )i o v net v
T T E E mc Q dxdydz c
t tV
x x ! p !
x x&& &
gdxdydz&
g&
v
T T T T k k k g c
x x y y z z tV
x x x x x x x !
x x x x x x x&
( )v
T T T T k k k dxdydz gdxdydz dxdydzc
x x y y z z tV
x x x x x x x !
x x x x x x x&
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2-3 Special forms of heat conduction equations
Constant thermal conductivity, k= constant
No heat generation source within the system
is called thermal diffusivity. It indicates how fast the heatdiffuses in the medium.
Steady state heat conduction with internal heat generation
Steady state and no internal heat generation in the medium
It is a Laplace equation
2 2 2
2 2 2
T T T g c T
x y z k k t
Vx x x x !
x x x x
&
t
T
t
T
k
c
z
T
y
T
x
T
x
x
x
x
x
x
x
x
x
x
E
V 12
2
2
2
2
2
ck VE /!
2 2 2
2 2 20
T T T g
x y z k
x x x !
x x x
&
022
2
2
2
2
2
!!xx
xx
xx T
zT
yT
xT
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2-3 One-dimensional heat conduction equations
One-dimensional (There is no
temperature gradient in y & z
directions), unsteady, constant kwith internal heat generation.
One-dimensional, steady state,
constant kwith internal heat
generation.
One-dimensional, steady state,
constant kand no internal heatgeneration.
2
20
d T g
dx k !
&
2
2
1T g T
x k tE
x x !
x x
&
2
20
d T
dx!
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2-4 The general heat conduction in cylindrical coordinates
Cylindrical coordinates with
constant k
One-dimensional (there is notemperature gradient in z and directions), unsteady, constant k,and with internal heat generation
One-dimensional, steady state,and constant kwith internal heat
generation
One-dimensional, steady state,constant k, and no internal heatgeneration.
10
d dT gr
r d r d r k !
&
0d d T
rdr dr
!
1
r
x
xTr
xT
xr
1
r2x
2T
xJ2
x2T
x2z
&g
k!
1
E
xT
xt
1 1T g Trr r r k tE
x x x !x x x
&
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2.4 Boundary conditions for steady state, one-dimensional heat conductions
Below is a plane wall with a thickness L. The left hand surface is located at x= 0 and the right hand side is located at x = L. The temperature or heat flux
at any point on the wall may be specified as boundary conditions. The
common boundary conditions for 1-dimensional, steady state hea
conduction problems are:
Constant surface temperature : If thetemperature at x = 0 is constant
T = T(0) = constant = To
Constant surface heat flux : if the heat flux across the plane x = 0 is
constant.
x
0 L
k(T
x
)x!0
! &q ! const.
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Adiabatic or insulated surface, there is no hat flow across theplane x = 0
Convection boundary condition at x = L, the ambient fluidtemperature is . The heat flow rate from the internal pointof the plate to the surface x = L is equal to the convection heattransfer rate to the ambient air
Interface boundary condition
0 L
Symmetric boundary condition
There is no heat flow across the symmetric plane.
0( ) 0
x
dT
dx!
!
( ) ( ) x L LT
k h T T
x
! g
x !
x
1 2
1 2
( ) ( )i i
i i
dT dT k k
dx dx
T T
!
!
1 2 q&
( ) 0sy etricT
x
x!
x
T
T
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Example 2-13 heat conduction in the base of an iron
Given : P = 1200W, L = 0.5cm, A = 300cm2, k = 15W/mK, = constant
= 20oC h = 80W/m2K
L = 0.5cm
Find - Temperature distribution x
- T(0) and T(0.5cm)
Assumption : the area is much larger than the thickness. One-dimensionapplication is possible, steady state operation, and constant k
Solution
- Governing equation and boundary conditions
q&
Tg
2
2
0 4 2 2
0
1200( )0, ( ) . 40,000( )
300 10 ( )
, ( ) [ ( ) ]
x
x L
d T
dxdT P W W
x q k constdx A x m m
dTx L k h T L T
dx
!
! g
!
! ! ! ! ! !
! !
&
A
q&T h
air
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-The solution : integrating once
Applying boundary condition 1
Integrating again
Applying boundary 2
The temperature profile
Temperature at x = 0
Temperature at x = 0.005m
1
dTc
dx
!
0 1
1
( ) 40,000
40,0002666.6( / )
15
x
o
dTk kc q
dx
c C m
! ! ! !
! !
&
1 1 2
1
2 1
[ ( ) ] [ ]
533.3o
kc h T L T h c L c T
kcc T c L C
h
g g
g
! !
! !
1 2T c x c!
2666.6 533
.
3T
x!
(0) 533.3oT C!
T(0.005) ! 2666.6x0.005 533.3o C ! 520o C
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Ex. 2-18 Variation of temperature in a resistance heater
Given : r o = 5mm, k = 100W/mK,
g = 5000W/m3
, To = 105o
C Find (1) temperature distribution
(2) maximum temperature
Assumptions
- steady state
- very long one-dimensional
- k = constant
- uniform internal heat generation
The governing equation & boundary conditions.
- Heat conduction equation
- the boundary conditions(1) constant surface T at r = ro
(2) symmetric Temperature
distribution at the center
ro
To= 105oC
10
d dT gr
r dr dr k
!&
0( ) 0rdT
dr!
!
T(r
0) ! T(5mm) ! 105o C
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- Integrating once
- Applying second boundary condition
- Rearranging the equation
- Integrating again
- Applying the first B. C.
- The temperature distribution
- The max. temperature occurs at x = 0
2
12
dT g
r r cdr k!
&
10; 0 0
dTr c
dr! ! !
2
dT gr
dr k!
&
r ! r
o;T ! T
o c
2!
&g
4kr
o
2 T
o
2
24
gT r ck
! &
T !
&g
4k(ro2
r
2
) T
o
Tmax
!&g
4kr
o
2 Tro
!5x108
4x100(0.005)2 105 ! 136.25
o C
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The following pages will not be taught
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2.5 Solution of steady one-dimensional heat conduction problem
(3) Heat conduction in a long solid circular cylinder with heat generation
Assumptions- One-dimensional and uniform internal heat generation
- k= constant
- Convection heat transfer coefficient is h and ambient fluid
temperature is T.
- Steady-state The 1-D heat conduction equation in cylindrical coordinates
The boundary conditions
The solution : Integrating once and apply the first boundary condition,
d dT gr r
dr dr k!
&
r ! 0 : dTdr
! 0,............r ! ro
: kdTdr
! h(Tr
o
Tg
)
2
12
dT gr r C
dr k!
&C1 = 0
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Repeat integration and get The temperature gradient at ro.
Applying the second boundary condition
Temperature at ro
The temperature profile is
or
The maximum temperature occurs at the center, r = 0.
2
24
gT r C
k!
& ( )2o
r odT g rdr k
! &
( ) ( ) ( )2o o o
r r o r
dT gk h T T r h T T
drg g
! ! &
2oo
rgrT T
hg
! &
2
2 04o
r
gC T r
k!
&
2 2( )4 o
o r
gT r r T
k!
&
2 2
0( )
4 2
ogrg
T r r Tk h
g!
&&