Chapters 2 Heat Conduction Pp

8/8/2019 Chapters 2 Heat Conduction Pp

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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!

&&

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Chapters 2 Heat Conduction Pp