Chapter 7 Heat Conduction

8/17/2019 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.

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Chapter 7 Heat Conduction