CUMT HEAT TRANSFER LECTURE Chapter 3 Steady-State Conduction Multiple Dimensions CHAPER 3 Steady-State Conduction Multiple Dimensions.

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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 20

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

20

XX

x

2

2

20

YY

y

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

This function cannot fit the sine-function boundary condition, so that the solution may be excluded.2 0


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25 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

This function cannot fit the sine-function boundary condition, so that the solution may be excluded.2 0


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29 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

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 C

9 11 120 y yC C e C e

9 10 11 120 cos sin y yC W C W C e C e

9 10 11 12sin cos sin H Hm

xT C x C x C e C e

W


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accordingly,

and from (c),

This requires that

11 12C C

9 0C

10 120 sin y yC C W e e

sin 0W


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then

which requires that Cn =0 for n >1.

We get

n

W

11

sin sinhnn

n x n yT T C

W W

1

sin sin sinhm nn

x n x n HT C

W W W

The final boundary condition may now be applied:


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The final solution is therefore

1

sinh /sin

sinh /m

y W xT T T

H W W

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 at m

y

x

x W

xT y H

W


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Using the first three boundary conditions, we obtain the solution in the form of Equation:

11

sin sinhnn

n x n yT T C

W W

Applying the fourth boundary condition gives

2 11

sin sinhnn

n x n HT T C

W W


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This series is

then

1

2 1 2 11

1 12sin

n

n

n xT T T T

n W

1

2 1

1 12 1

sinh /

n

nC T Tn H W n

The final solution is expressed as

1

1

12 1

1 1 sinh /2sin

sinh /

n

n

n y WT T n x

T T n W n H W


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0 at 0

0 at 0

0 at

sin at m

y

x

x W

xT y H

W

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.

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Note that the inverse hyperbolic cosine can be calculated from

1 2cosh 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

AS

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 6m n m nm n m n m n

T x T x TT T x

x x x

2 2 3 3

1, , 2 3, , ,

( ) ( )...

2 6m n m nm n m n m n

T x T x TT T x

x x x


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22

1, 1, , 2

,

2 ( ) ...m n m n m n

m n

TT T T x

x

21, , 1, 2

2 2

,

2( )

( )m n m n m n

m n

T T TTo x

x x

22

1,,1,

,

2

2

)()(

2yo

y

TTT

y

T nmnmnm

nm


Differential equation for two-dimensional steady-state heat flow

2 2

2 20

T T q

x y k


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Discrete equation at nodal point (m,n)

1, , 1, , 1 , , 1

2 2

2 20m 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 20m 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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2 、 Discrete equationThermal balance

(1) Interior pointssteady-state & no heat

generation

1, 1, , 1 , 1 0m n m n m n m nq q q q

1, ,1,

d

dm n m n

m n

T TTq kA k y

x x

x

TTykq nmnm

nm

,,1,1

, 1 ,, 1

m n m nm n

T Tq k x

y

, 1 ,

, 1m n m n

m n

T Tq 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 Tk 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

21, 1, , 1 , 1 ,4 ( ) 0m n m n m n m n m n

qT T T T T x

k

steady-state with heat generation

(1) Interior points


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Thermal balance(2) boundary

points1, ,

1,m n m n

m n

T Tq k y

x

, 1 ,, 1 2

m n m nm n

T Txq k

y

, 1 ,, 1 2

m n m nm n

T Txq k

y

,( )w m nq h y T T

1, , , 1 , , 1 ,,( )

2 2m n m n m n m n m n m n

m n

T T T T T Tx xk 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) ( )

2m n m n m n m n

h x h xT T T T T

k k

1, , , 1 ,, ,( ) ( )

2 2 2 2m n m n m n m n

m n m n

T T T Ty x x yk k h T T h T T

x y

, 1, , 1

1( 1) ( )

2m n m n m n

h x h xT 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 nm n m n

T T T T T Ty xk y k k

x x y

T T x yk x h T T h T T

y

, 1, 1, , 1 , 1

1( 3) (2 2 )

2m n m n m n m n m n

h x h xT 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 aA

a a a

1

2

...

n

T

TT

T

1

2

...

n

C

CC

C

A T CIteration 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 11 ( ) 100

3 2 3T T T

3 2 6

6 3 5 9

7 4 8

8 7 5 9

1 1 12 (500 2 ) 100

3 2 31 1 1

2 ( 2 ) 1003 2 31 1 1

2 (100 2 ) 1003 2 31 1 1

2 ( 2 ) 1003 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 12 (500 2 ) 100

3 2 3T T T

6 3 5 9

1 1 12 ( 2 ) 100

3 2 3T T T T

7 4 8

1 1 12 (100 2 ) 100

3 2 3T T T

8 7 5 9

1 1 12 ( 2 ) 100

3 2 3T T T T

The equation for node 9 is

9 6 8

1 1 11 ( ) 100

3 2 3T 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 xq k A x x

y y y y

W m

The heat flow out of the 100 face is℃

71 41

( 100)( 100) ( 100)10 [ ]

2... 3019 /

TT TT yq k A y y

x x x xW m

Example 3-5

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2

3 6 9

( )

10 [ ( 100) ( 100) ( 100)]2

... 1214.6 /

q h A T T

yy 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

yy 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 ii

j i j

T Tq

R

qi — heat generation, radiation, etc.i — solving nodej — 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

xR

kA

1m m n nR R R R

k


0j ii

j i j

T Tq

R

so

3-6 Numerical Formulation in Terms of Resistance Elements

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3-7 Gauss-Seidel Iteration

0j ii

j i j

T Tq

R

( / )

(1/ )

i j i jj

ii j

j

q T R

TR

StepsAssumed initial set of values for Ti；Calculated Ti according to the equation；

—using the most recent values of the TiRepeated the process until converged.

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Convergence Criterion

( 1) ( )i n i nT T ( 1) ( )

( )

i n i n

i n

T T

T

3 610 10 ～

Biot number

h xBi

k

3-7 Gauss-Seidel Iteration

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Example 3-6

1xR

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 jj j j

ii j j j

j j j

q T R q k T k T

TR k k

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Example 3-6

Because each node has four resistance connected to it and k is assumed constant, so

4jj

k k 1

4i jj

T T

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3-8 Accuracy Consideration

Truncation Error — Influenced by difference schemeDiscrete Error — Influenced by truncation error & x△Round-off Error — Influenced by x△

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Summary

(1)Numerical MethodSolving ZoneNodal equationsthermal balance method — Interior & boundary pointAlgebraic equationsGauss-Seidel iteration

( / )

(1/ )

i j i jj

ii j

j

q T R

TR

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Summary

(2)Resistance Forms

0j ii

j i j

T Tq

R

(3)Convergence

Convergence Criterion

( 1) ( )i n i nT T

( 1) ( )i n i nT T

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Summary

(4)Accuracy

Truncation ErrorDiscrete ErrorRound-off Error

Important conceptionsNodal equations — thermal balance methodCalculated temperature & heat flowConvergence criterionHow to improve accuracy

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Exercises

Exercises: 3-16, 3-24, 3-48, 3-59

CUMT HEAT TRANSFER LECTURE Chapter 3 Steady-State Conduction Multiple Dimensions CHAPER 3 Steady-State Conduction Multiple Dimensions.

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