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Two-Dimensional Conduction:

Finite-Difference Equations

and

Solutions

Chapter 4

Sections 4.4 and 4.5

Numerical methods

analytical solutions that allow for the determination of the exact

temperature distribution are only available for limited ideal cases.

graphical solutions have been used to gain an insight into complex heat

transfer problems, where analytical solutions are not available, but they

have limited accuracy and are primarily used for two-dimensional

problems.

advances in numerical computing now allow for complex heat transfer

problems to be solved rapidly on computers, i.e., "numerical techniques.

current numerical techniques include: finite-difference analysis; finite

element analysis (FEA); and finite-volume analysis.

in general, these techniques are routinely used to solve problems in heat

transfer, fluid dynamics, stress analysis, electrostatics and magnetics, etc.

We will show the use of finite-difference analysis to solve conduction heat

transfer problems.

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Finite-difference Analysis

numerical techniques result in an approximate solution, however the error

can be made very small.

properties (e.g., temperature) are determined at discretepoints in the region

of interest-these are referred to as nodal points or nodes.

Consider the finite-difference technique for 2-D conduction heat transfer:

in this case each node represents the temperature of a point on the surface

being considered.

the temperature at the node represents the average temperature of that

region of the surface.

algebraic expressions are used to define the relationship between adjacent

nodes on the surface usually the boundary conditions are specified.

by increasing the number of nodes on the surface being considered it is

possible to increase the spatial resolution of the solution and to potentially

increase the accuracy of the numerical solution, however this increases thenumber of calculation is required to obtain a solution to the problem.

Finite-Difference Approximation

The Nodal Network and Finite-Difference Approximation The nodal networkidentifies discrete points at which the temperature is to be

determined and uses an m,n notation to designate their location.

What is represented by the temperature determined at a nodal point,

as for example, Tm,n?

How is the accuracy of the solution affected by construction of the nodal

network?

What are the trade-offs between selection of afine or a coarse mesh?

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Finite-Difference Method

The Finite-Difference Method

Procedure:

Represent the physical system by a nodal network i.e.,

discretization of problem.

Use the energy balance method to obtain a finite-difference

equation for each node of unknown temperature.

Solve the resulting set of algebraic equations for the unknown

nodal temperatures.

Use the temperature field and Fouriers Law to determine theheat transfer in the medium

Finite difference formulation of the differential equation

numerical methods are used for solving differential equations,

i.e., the DE is replaced by algebraic equations

in the finite difference method, derivatives are replaced by

differences, i.e.,

this is based on the premise that a reasonably accurate result

can be obtained by replacing differential quantities by

sufficiently small differences, letting

( ) ( ) ( )d f x f x x f x

dx x

+

0

( ) ( )limx

d f x f x

dx x

=

x

x

f( )x x+

( )x

( )x

x x+ x

Tangent line

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Finite-Difference Approximation

Finite-Difference Formulation of Differential Equation

For example: Consider the 1-D steady-state heat conduction equation with internal

heat generation) i.e.,

For a pointm,n we approximate the first derivatives at points m-x and m+ x as

2

20

T q

kx

+ =

x


example: 1-D steady-state heat conduction equation with internal heat generation

For a pointm we approximate the 2nd derivative as

Now the finite-difference approximation of the heat conduction equation is

This is repeated for all the modes in the region considered

1122

1 12

2

1 12

2

d T d T m m m mdx dx mm

m

m m m

T T T T

T x x

x xx

T T Tx

+

+

+

+

1 12

20m m m m

T T T q

kx

+ + + =

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If this was a 2-D problem we could also construct a similar relationship in the

both the x and Y-direction at a point (m,n) i.e.,

Now the finite-difference approximation of the 2-D heat conduction equation is

Once again this is repeated for all the modes in the region considered. We could also

derive a similar equation for the 3-D case

( )

2, 1 , , 1

2 2

,

2m n m n m n

m n

T T TT

y y

+ +

( )

21, , 1,

2 2

,

2m n m n m n

m n

T T TT

x x

+ +

( ) ( )

1, , 1, , 1 , , 1

2 2

2 20

m n m n m n m n m n m nT T T T T T q

kx y

+ + + ++ + =

Finite-Difference Formulation of Differential EquationIf x =y, then the finite-difference approximation of the 2-D heat conduction equation is

which can be reduced to

and the relationship reduces to

if there is no internal heat generation,

Which is just the average of the surrounding nodes temperatures!

( )2

1, , 1, , 1 , , 12 2 0m n m n m n m n m n m nq x

T T T T T T k

+ +

+ + + + =

( )2

1, 1, , 1 , 1 ,4 0m n m n m n m n m nq x

T T T T T k

+ +

+ + + + =

( )2

, 1, 1, , 1 , 1

1

4m n m n m n m n m n

q xT T T T T

k + +

= + + + +

, 1, 1, , 1 , 1

1

4m n m n m n m n m nT T T T T + + = + + +

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Consider this simple case

Consider this simple case

100 100 100

50 200

50 200

50 200

300 300 300

100 100 100

50 200

50 200

50 200

300 300 300

100 100 100

50 200

50 200

50 200

300 300 300

[ ]1 3 21

100 504

T T T= + + +

[ ]2 1 41

100 2004

T T T= + + +

[ ]3 1 41

300 504

T T T= + + +

[ ]4 3 21

300 2004

T T T= + + +

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A summary of finite-difference equations for common nodal regions is provided

in Table 4.2.

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Energy Balance Method (cont.)

Consider an external corner with convection heat transfer.

( ) ( ) ( ) ( ) ( ) ( )1, , , 1 , , 0m n m n m n m n m nq q q + + =

( ) ( )

1, , , 1 ,

, ,

2 2

x yh h 0

2 2

m n m n m n m n

m n m n

T T T T y xk k

x y

T T T T

+

+ + =

or, with ,x y = 1, , 1 ,

h h2 2 1 0m n m n m n

x xT T T T

k k

+ + + =

(4.43)

e_04_02

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p_04_45

fig_04_06

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p_04_44

e_04_04_02

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Energy Balance Method (cont.)

Note potential utility of using thermal resistance concepts to express rate

equations. E.g., conduction between adjoining dissimilar materials with

an interfacial contact resistance.

( ) ( ), 1 ,

, 1 ,

m n m n

m n m ntot

T Tq

R

=

( ) ( ),/ 2 / 2t c

tot

A B

Ry yRk x x k x

= + +

(4.46)

p_04_38

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Solution Methods

Solutions Methods Matrix Inversion: Expression of system of N finite-difference equations for

N unknown nodal temperatures as:

[ ][ ] [ ]A T C= (4.48)

Coefficient

Matrix (NxN)Solution Vector

(T1,T2, TN)Right-hand Side Vector of Constants

(C1,C2CN)

Solution [ ] [ ] [ ]1T A C

=

Inverse of Coefficient Matrix (4.49)

Gauss-Seidel Iteration: Each finite-difference equation is written in explicit

form, such that its unknown nodal temperature appears alone on the left-

hand side:( ) ( )1 ( 1)

1 1

i Nij ijk k kii j j

j j iii ii ii

a aCT T T

a a a

= = +

= (4.51)

where i =1, 2,, N andk is the level of iteration.

Iteration proceeds until satisfactory convergence is achieved for all nodes:

( ) ( )1k ki iT T

What measures may be taken to insure that the results of a finite-difference

solution provide an accurate prediction of the temperature field?

Problem: Finite-Difference Equations

Problem 4.39: Finite-difference equations for (a) nodal point on a diagonalsurface and (b) tip of a cutting tool.

(a) Diagonal surface (b) Cutting tool.

Schematic:

ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties

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Problem: Finite-Difference Equations (cont.)

ANALYSIS: (a) The control volume about node m,n is triangular with sides x and y and diagonal

(surface) of length 2 x.

The heat rates associated with the control volume are due to conduction, q1and q2, and to convection,

qc. An energy balance for a unit depth normal to the page yields

inE 0=

1 2 cq q q 0+ + =

( ) ( ) ( )( )

m,n-1 m,n m+1,n m,n

m,n

T T T Tk x 1 k y 1 h 2 x 1 T T 0.

y x

+ + =

With x = y, it follows that

m,n-1 m+1,n m,nh x h x

T T 2 T 2 2 T 0.k k

+ + + =

(b) The control volume about node m,n is triangular with sides x/2 and y/2 and a lower diagonal

surface of length ( )2 x/2 .

The heat rates associated with the control volume are due to the uniform heat flux, qa, conduction, qb,

and convection qc. An energy balance for a unit depth yields

ina b c

E =0q q q 0+ + =

( )m+1,n m,n

o m,n

T Tx y x

q 1 k 1 h 2 T T 0.2 2 x 2

+ + =

or, with x = y,

m+1,n o m,nh x x h x

T 2 T q 1 2 T 0.k k k

+ + + =

Problem: Cold Plate

Problem 4.76: Analysis of cold plate used to thermally control IBM multi-chip,

thermal conduction module.

Features:

Heat dissipated in the chips is transferred

by conduction through spring-loaded

aluminum pistons to an aluminum cold

plate.

Nominal operating conditions may be

assumed to provide a uniformly

distributed heat flux of

at the base of the cold plate.

5 210 W/mo

q=

Heat is transferred from the cold

plate by water flowing through

channels in the cold plate.

Find: (a) Cold plate temperature distribution

for the prescribed conditions. (b) Options

for operating at larger power levels while

remaining within a maximum cold plate

temperature of 40C.

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Problem: Cold Plate (cont.)

Schematic:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant properties.


ANALYSIS:

Finite-difference equations must be obtained for each of the 28 nodes. Applying the energybalance method to regions 1 and 5, which are similar, it follows that

ode 1: ( ) ( ) ( ) ( )2 6 1 0y x T x y T y x x y T + + =

ode 5:( ) ( ) ( ) ( )4 10 5 0y x T x y T y x x y T + + =

Nodal regions 2, 3 and 4 are similar, and the energy balance method yields a finite-difference equation of

the form

Nodes 2,3,4:

( )( ) ( ) ( ) ( )1, 1, , 1 ,2 2 0m n m n m n m ny x T T x y T y x x y T + + + + =

Energy balances applied to the remaining combinations of similar nodes yield the following finite-difference

equations.

odes 6, 14: ( ) ( ) ( ) ( ) ( )[ ] ( )1 7 6x y T y x T x y y x h x k T h x k T + + + =

( ) ( ) ( ) ( ) ( )[ ] ( )19 15 14x y T y x T x y y x h x k T h x k T + + + =

odes 7, 15: ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )6 8 2 72 2 2x T T x y T y x x y h x k T h x k T + + + + =

( )( ) ( ) ( ) ( ) ( )[ ] ( )14 16 20 152 2 2y x T T x y T y x x y h x k T h x k T + + + + =

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odes 8, 16: ( ) ( ) ( ) ( ) [ ( ) ( )7 9 11 32 2 3 3x T y x T x y T x y T y x x y + + + +

( )( )] ( ) ( )8h k x y T h k x y T + + = +

( ) ( ) ( ) ( ) [ ( ) ( )15 17 11 212 2 3 3x T y x T x y T x y T y x x y + + + +

( )( )] ( ) ( )16h k x y T h k x y T + + = +

ode 11: ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )8 16 12 11x y T x y T 2 y x T 2 x y y x h y k T 2h y k T + + + + =

odes 9, 12, 17, 20, 21, 22:( ) ( ) ( ) ( ) ( ) ( )[ ]1, 1, , 1 , 1 ,2 0m n m n m n m n m ny x T y x T x y T x y T x y y x T + + + + + + =

odes 10, 13, 18, 23:

( ) ( ) ( ) ( ) ( )[ ]1, 1, 1, ,2 2 0n m n m m n m nx y T x y T y x T x y y x T+ + + + =

Node 19: ( ) ( ) ( ) ( ) ( )[ ]14 24 20 192 2 0x y T x y T y x T x y y x T + + + =

Nodes 24, 28: ( ) ( ) ( ) ( )[ ] ( )19 25 24 ox y T y x T x y y x T q x k + + =

( ) ( ) ( ) ( )[ ] ( )23 27 28 ox y T y x T x y y x T q x k + + =

odes 25, 26, 27:

( ) ( ) ( ) ( ) ( )[ ] ( )1, 1, , 1 ,2 2 2m n m n m n m n ox T y x T x y T x y y x T q x k + + + + + =


Evaluating the coefficients and solving the equations simultaneously, the steady-state temperaturedistribution (C), tabulated according to the node locations, is:

23.77 23.91 24.27 24.61 24.74

23.41 23.62 24.31 24.89 25.07

25.70 26.18 26.33

28.90 28.76 28.26 28.32 28.35

30.72 30.67 30.57 30.53 30.52

32.77 32.74 32.69 32.66 32.65

(b) For the prescribed conditions, the maximum allowable temperature (T24= 40C) is reached when

oq = 1.407 105W/m2(14.07 W/cm2).

Options for extending this limit could include use of a copper cold plate (k 400 W/mK) and/or increasing

the convection coefficient associated with the coolant.

With k = 400 W/mK, a value of oq = 17.37 W/cm2may be maintained.

. With k = 400 W/mK and h = 10,000 W/m2K (a practical upper limit), oq = 28.65 W/cm

2.

Additional, albeit small, improvements may be realized by relocating the coolant channels closer to the base

of the cold plate.

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