Comp Heat Transfer4 Repaired)

7/31/2019 Comp Heat Transfer4 Repaired)

1/48

Chapter Four Finite Different Application to Steady State Heat Conduction

Chapter Four FINITE DIFFERENT APPLICATION TO STEADY STATE HEAT

CONDUCTION


2/48


In this chapter, the modeling of steady state heat conduction problems by the finite

difference technique is discussed. Several illustrative examples are presented in rectangular

and cylindrical coordinates system.

4.1 Steady State Heat Conduction

We begin with the simplest case as shown in Fig. 4.1. Here we only consider a steady state

one-dimensional conduction mode of heat transfer without heat generation. Therefore, the

heat conduction equation reduces to

02

2

=x

T(4.1)

We discretize Eq. (4.1) using the finite difference approximation and gives

( )0

22

11 =

+ +x

TTT iii(4.2)

or

2

11 + += iii TTT (4.3)

Note that the central difference with second order accuracy was used to approximate Eq.

(4.1).

Figure 4.1 Heated steel bar


3/48


By referring to Fig. 4.1, we evaluate the temperature at each nodes using Eq. (4.2) with the

following boundary condition

5001

=T (4.4)

1004 =T (4.5)

and

2

500

2

3132

+=

+=

TTTT (4.6)

2

100

2

2243

TTTT

+=

+= (4.7)

Equations (4.6) and (4.7) can be easily solved and get

2

3

366.67

233.33

T

T

== (4.8)

Next, we consider steady state with no heat generation, two-dimensional heat flow equation

with the following governing equations

02

2

2

2

=

+

y

T

x

T(4.9)

Thus, the finite difference approximation for the above equation becomes

( ) ( )0

222

,1,1,

2

,,1,1 =

++

+ ++y

TTT

x

TTT jijijijijiji(4.10)

If yx = then

04 ,1,1,,1,1 =+++ ++ jijijijiji TTTTT (4.11)

or


4/48


4

1,1,,1,1

,

++ +++= jijijijijiTTTT

T (4.12)

Figure 4.2 Four nodes problem

Another simple example is shown in Figure 4.2 and the four equations for nodes 6, 7, 10

and 11 would be

Node 6

4

500100 1076

TTT

+++= (4.13)

Node 7

4

500100 1167

TTT

+++= (4.14)

Node 10

4

100100 61110

+++=

TTT (4.15)

Node 11


5/48


4

100100 71011

+++=

TTT (4.16)

Since symmetrical boundary condition for the left and right sides

1110

76

TT

TT

==

(4.17)

and the final results are

150

250

1110

76

====

TT

TT(4.18)

4.2 Iterative Methods

For the number of nodes very large, any iterative technique may frequently yield a more

efficient solution. The basis of all iterative methods is that an approximate solution is

guessed for each point at which solution is sought and the values are then repeatedly

updated. One such method is called aJacobi method

. The procedure in Jacobi method is as

follow

1. An initial set of values for the Tis assumed. For a large number of nodes,

the Tare usually assigned a zero value to start the calculation

2. The new value of the nodal temperature are calculated

3. The process is repeated until successive calculation differs by a sufficient

small amount.

For practice purpose, we return to the first case above and the updating equation for

temperature is


6/48


2

50032

+=T

T (4.19)

2

100 23

TT

+= (4.20)

Tab. 4.1 shows the solution for the current problem using the Jacobi method

Table 4.1 Solution by Jacobi iteration method

Number of iteration2T 3T

0 0 0

1 250 50

2 275 1753 337.5 187.5

4 343.75 218.75

5 359.375 221.875

6 360.9375 229.6875

7 364.8438 230.4688

8 365.2344 232.4219

18 366.6653 233.3324

19 366.6662 233.3326

20 366.6667 233.3333

The compute program code written in MATLAB is shown below

%One-dimensional conduction heat transfer without heat generation

%Jacobi method

clear all; clc; close all;

lmax=100;

imax=4;

imm1=imax-1;

t = zeros;

t(1)=500;


7/48


t(imax)=100;

for l =1:lmax

for i=2:imm1

tnew(i)=(t(i+1)+t(i-1))/2;

end

for i=2:imm1

t(i)=tnew(i);

end

end

Tab. 4.2 shows the predicted value of temperature for the case of two-dimensional problem

using the discussed Jacobi iteration method.

Table 4.2 Solution by Jacobi iteration method for two-dimensional problem

Number of iteration 6T 7T 10T 11T

0 0 0 0 0

1 150 150 50 50

2 200 200 100 100

3 225 225 125 125

4 237.5 237.5 137.5 137.5

5 243.75 243.75 143.75 143.75

6 246.875 246.875 146.875 146.875

After six iterations, the approximate solution is


8/48


)150(875.146

)150(875.146

)250(875.246

)250(875.246

11

10

7

6

====

T

T

T

T

(4.21)

with the exact solution shown in parentheses for comparison. More iteration will take us

even nearer to this exact solution.


%Two-dimensional conduction heat transfer without heat generation

%Jacobi method


lmax=100;

imax=4;

jmax=4;

imm1=imax-1;

jmm1=jmax-1;

t = zeros;

for i=1:imax

t(i,jmax)=500;

t(i,1)=100;

end

for j=1:jmax

t(1,j)=100;

t(imax,j)=100;

end

for l =1:lmax


9/48


for i=2:imm1

for j=2:jmm1

tnew(i,j)=(t(i+1,j)+t(i-1,j)+t(i,j+1)+t(i,j-1))/4;

end

end

for i=2:imm1

for j=2:jmm1

t(i,j)=tnew(i,j);

end

end

end

Another well-known technique is the Gauss-Seidel iterative method. This method is very

similar to the Jacobi method. The sole different is that, when sequentially iterating with the

update equations, new updated value are used as soon as they are available. Example

calculation for two-dimensional case is shown as follow

4

500100 1076

TTT

+++= (4.16)

4

500100 1167

TTT

+++= (4.17)

4

100100 61110

+++=

TTT (4.18)


10/48


4

100100 71011

+++=

TTT (4.19)

Table 4.2 Solution by Gauss-Seidel iteration method

Number of iteration 6T 7T 10T 11T

0 0 0 0 0

1 150 187.5 87.5 118.75

2 218.75 234.375 134.375 142.18

3 242.18 246.09 146.09 148.05

4 248.05 249.025 149.025 149.51

They can be seen that, after only four iterations, the approximate solution is

)150(51.149

)150(025.149

)250(025.249

)250(05.248

11

10

7

6

====

T

T

T

T

(4.20)


%Two-dimensional conduction heat transfer without heat generation


lmax=100;

imax=4;

jmax=4;

imm1=imax-1;

jmm1=jmax-1;

t = zeros;

for i=1:imax

t(i,jmax)=500;

t(i,1)=100;

end

for j=1:jmax


11/48


t(1,j)=100;

t(imax,j)=100;

end

for l =1:lmax

for i=2:imm1

for j=2:jmm1

tnew(i,j)=(t(i+1,j)+t(i-1,j)+t(i,j+1)+t(i,j-1))/4;

t(i,j)=tnew(i,j);

end

end

end

4.2.1 Boundary treatments

When the solid is exposed to some other type of boundary conditions; i.e convection

boundary condition, the temperature at the surface must be computed differently from the

method given above. To see this, now, let consider a long rectangular bar as shown in Fig.

4.3.


12/48


Figure 4.3 Two dimensional heat conduction in a bar (Physical problem)

In the absence of heat source, the heat conduction equation for this geometry is given by

02

2

2

2

=

+

y

T

x

T(4.21)

The following boundary conditions are applied on the side surfaces of the bar:

0TT= atx = 0 (4.22)

0=y

Taty = 0 (4.23)

qx

Tk =

atx =L (4.24)

( )fTThxTk = aty =H (4.25)

Using central difference approximation for the second derivatives, the nodal equation at

point (i,j) is given by


13/48


14/48


Figure 4.4 Two dimensional heat conduction in a bar (grid)

The prescribed temperature condition at i = 1 becomes

0,1 TT j = for 11 + Mj (4.28)

On the bottom surface ( 1=j ), the discretised heat conduction equation is determined as

04 1,0,2,1,11,1 =+++ + iiiii TTTTT (4.29)

Here, it is required to introduce an imaginary node point 0,iT at a distance y from the

bottom surface. We know that the insulated boundary can be discretized as

02

0,2,

1,

=

=

y

TT

y

T ii

i

(4.30)

or


15/48


0,2, ii TT = (4.31)

and we obtain

042 1,2,1,11,1 =++ + iiii TTTT for Ni 2 (4.32)

And therefore

4

2 2,1,11,11,

iii

i

TTTT

++= + for Ni 2 (4.33)

For the right boundary, the heat flux condition is applied as follow. Consider an imaginary

point outside the boundary, then the discretised form of the boundary condition becomes

qx

TTk

x

Tk

jNjN =

=

+2

,,2(4.34)

or

jNjN Tk

xqT ,,2

2+

=+ (4.35)

Combine the above equation for the heat conduction equation for the right boundary node

gives

k

xqTTTT jNjNjNjN

=++ ++++

242 ,11,11,1, for Mj 2 (4.36)

and therefore

jN

jNjNjN

jN Tk

xqTTT

T ,1

1,11,1,

,1 44

22

+

+++

+

++

=for Mj 2 (4.37)

In a similar manner, the nodal equations for the top boundary can be obtained by

discretizing the convective condition using imaginary point


16/48


( )fTTh

y

Tk =

(4.38)

or

( )f

MiMiTT

k

h

y

TT=

+

2

,2,(4.39)

and

( )fMiMi TT

k

yhTT

=+

2,2, (4.40)

Substituting for this imaginary point temperature in the heat conduction equation give the

nodal equations as follow

fMiMiMiMi Tk

yhT

k

yhTTT

=

+++ ++++ 2222 1,,1,11,1 for Ni 2 (4.41)

Next, let consider the corner nodes of the domain. The nodal equations of the corner points

on the prescribed temperature boundary are

01,1

01,1

TT

TT

M =

=

+ (4.42)

For the corner node ( 1,1+N ), two imaginary points ( 0,1+N ) and ( 1,2+N ) have to be

considered. The boundary conditions on the lower and the right surfaces are discretised

using these imaginary points. After combining the discretised boundary condition with the

heat conduction equation, the following nodal equation is obtained

kxqTTT NNN =+ ++ 1,12,11, 2 (4.43)

Similarly, by incorporating the boundary condition of heat flux and convective heat

exchange with the help of two imaginary points, the equation for the node ( 1,1 ++ MN ) is

obtained as


17/48


k

xqT

k

yhT

k

yhTT fMNMNMN

+

=

++ ++++ 1,1,11, 2 (4.44)

The other nodal point equation for corner boundary nodes exposed to ambient temperature

fT is shown below

Bi1

Bi2,11,1,1 +

++= ++++

fNMNM

NM

TTTT

wherek

xh=Bi

Bi3

2Bi 1,,1,11,, +

+++++= ++ jijijijifji

TTTTTT

wherek

xh=Bi

Figure 4.6 Convection boundary nodes

Example 4.1

A 1-by 2-cm ceramic strip (k= 3W/moC, =1600kg/m3 and c = 0.8kJ/kgoC) is embedded

in a high-thermal-conductivity material as shown in Fig. 4.7 so that the sides are

maintained at a constant temperature of 900oC. The bottom surface of the ceramic is


18/48


insulated, and the top surface is exposed to a convection environment at T= 50oC, h =

50W/m2. Solve the steady state temperature distribution of the nodes.

Figure 4.7 Heated ceramic strip

Solution

Since the problem is steady state heat conduction without heat generation, the governing

equation is expressed as

02

2

2

2

=

+

y

T

x

T(4.45)

and the discretised equation is

( ) ( )0

222

,1,1,

2

,,1,1 =

++

+ ++y

TTT

x


Since yx = and therefore


or


19/48


4

1,1,,1,1

,


T (4.48)

Let us employ a uniform square grid as shown in Figure below.

Fig. 4.8 Uniform square grid for heated ceramic

For the bottom boundary, the adiabatic boundary condition is applied as follow. Consider

an imaginary point outside the boundary, then the discretised form of the boundary

condition becomes

042 1,2,1,11,1 =++ + iiii TTTT (4.49)

or

4

22,1,11,1

1,

iii

i

TTTT

++=

+

for 3,2,1=i and 1=j (4.50)

In a similar manner, the nodal equations for the top boundary can be obtained by

discretizing the convective condition using imaginary points and substituting for the

imaginary point temperature in the heat conduction equation. The resulting nodal equations

are


20/48


fMiMiMiMi Tk

yhT

k

yhTTT

=

+++ ++++ 2222 1,,1,11,1 or (4.51)

+

+++

=+++

+

k

yh

TTTTk

yh

TMiMiMif

Mi

22

22 ,1,11,11, for 3,2,1=i and 1+= Mj (4.52)

The computer program code written in MATLAB is shown below

clear all;clc;close all;

width = 0.02;

height = 0.01;

h = 50;

k = 3;

t_air = 50;

t_wall = 900;

l=0; %

lmax=100;

imax=5;

jmax=3;

imm1=imax-1;

jmm1=jmax-1;

dx = width/imm1;

dy = height/jmm1;

for j = 1:3

t(1,j) = t_wall;

t(imax,j) = t_wall;

end


21/48


for l=1:100

for i=2:imm1

j = 1;

tnew=(t(i+1,j)+t(i-1,j)+2.*t(i,j+1))/4;

t(i,j)=tnew;

end

for i=2:imm1

j = 2;

tnew=(t(i+1,j)+t(i,j+1)+t(i,j-1)+t(i-1,j))/4;

t(i,j)=tnew;

end

for i=2:imm1

j = jmax;

tnew=((2*h*dy/k).*t_air + t(i-1,j)+t(i+1,j)+2.*t(i,j-1))/

(2*(2+h*dy/k));

t(i,j)=tnew;

end

end

The value of temperature at all nodes after 100 iteration is shown in Table 4.2.

Table 4.2 Temperature distribution for Example 4.1

No. of Node Temperature No. of Node Temperature

1 822.645 5 843.8972

2 801.9496 6 848.7429


22/48


3 822.6465 7 868.4098

4 858.7429 8 856.1535

9 868.4098

Example 4.2

Consider the square as shown in Fig. 4.9. The left surface is maintained at 100oC and the

top surface 500oC, while the other two surfaces are exposed to an environment at 100oC.

The block is 1m square. Compute the temperature of the various nodes in the figure.

CmWh

2

10= andmCWk 10

=

Solution



02

2

2

2

=

+

y

T

x

T(4.53)


( ) ( )0

222

,1,1,

2

,,1,1 =

++

+ ++y

TTT

x




Therefore, for the internal nodes the temperature can be calculated by

4

1,1,,1,1

,


T (4.56)


23/48


Figure 4.9 Heated square block

Solution

For the bottom boundary nodes (j =1), since they are exposed to ambient temperature, the

general equation can be written as

04 1021111 =+++ + i,i,i,,i,i TTTTT (4.57)

From the convective boundary condition, we get

( )fi

i,i,TTh

y

TTk =

1,20

2(4.58)

This leads to

( ) 2,1,02

ifii, TTTk

yhT +

= (4.59)

Substitute into Eq. (4.58) yields


24/48


02

42

2 1,21111 =

+

+

++ + fii,,i,i T

k

yhT

k

yhTTT (4.60)

or

+

+++

=+

42

22 21111

1,

k

yh

Tk

yhTTT

Tfi,,i,i

i (4.61)

Since the right surface is also exposed to ambient temperature, the equation for the right

boundary nodes (i = imax) can be expressed as

04max1max1max1max1max =+++ ++ ,ji,ji,ji,ji,ji

TTTTT(4.62)

where

( )fji

,jijiTTh

x

TTk =

+ max,

1max,1max

2(4.63)

This leads to

( )jifji,ji TTT

k

xhT ,1maxmax,1max

2+ +

= (4.64)


022

42

1max1max,1maxmax, =+++

+

+

+ ,ji,jijifji TTTT

k

xhT

k

xh

(4.65)

or

+

+++= +

42

2

2 1max1max1maxmax,

k

xhTk

xh

TTTTf,ji,ji,ji

ji (4.66)

Now, we demonstrate the treatment for the bottom right corner node (imax,1).

The discretized equation for this node is expressed as


25/48


04 1max0max2max11max11max =+++ + ,i,i,i,i,i TTTTT (4.67)

Here, we need to replace two imaginary nodes (imax+1,1) and (imax, 0). Since the right

and bottom boundary are exposed to the ambient temperature, therefore

( )2max,1max,0max

2ifi,i TTT

k

yhT +

= (4.68)

and

( ) 1,1max1max,11max2

+ +

= ifi,i TTTk

xhT (4.69)

Substitute Eq. (4.69) and (4.68) into Eq. (4.67) gives

02222

422

2max1,1max1max, =++

+

+

+

+

,iifi TTT

k

yh

k

xhT

k

yh

k

xh(4.70)

Then

+

+

++

+

=

422

2222

2max1,1max

1max,

k

yh

k

xh

TTTk

yh

k

xh

T,iif

i (4.71)



width = 1.00;

height = 1.00;

h = 10;

k = 10;

t_air = 100;

t_left_wall = 100;

t_top_wall = 500;

l=0; %


26/48


lmax=100;

imax=5;

jmax=5;

imm1=imax-1;

jmm1=jmax-1;

dx = width/imm1;

dy = height/jmm1;

for j = 1:jmax

t(1,j) = t_left_wall;

end

for i = 1:imax

t(i,jmax) = t_top_wall;

end

for l=1:100

for i=2:imm1

for j=2:jmm1

tnew=(t(i+1,j)+t(i-1,j)+t(i,j+1)+t(i,j-1))/4;

t(i,j)=tnew;

end

end

for i=2:imm1

j = 1;

tnew=(t(i+1,j)+t(i-1,j)+2*t(i,j+1)+(2*dy*h*t_air)/k)/

(((2*dy*h)/k)+4);


27/48


t(i,j)=tnew;

end

for j=2:jmm1

i = imax;

tnew=(t(i,j+1)+t(i,j-1)+2*t(i-1,j)+(2*dx*h*t_air)/k)/

(((2*dx*h)/k)+4);

t(i,j)=tnew;

end

j=1;

i=imax;

tnew=((2*dx*h*t_air)/k + (2*dy*h*t_air)/k +2*t(i-1,j)

+2*t(i,j+1))/((2*dx*h)/k + (2*dy*h)/k+4);

t(i,j)=tnew;

end

Example 4.3

The fin shown in Fig. 4.10 has a base maintained at 300oC and is exposed to the convective

environment indicated. Calculate the steady state temperature of the nodes shown in the

figure ifk= 1.0W/moC and h = 40W/m2oC


28/48


Figure 4.10 Heated fin exposed to ambient temperature



02

2

2

2

=

+

y

T

x

T(4.72)


( ) ( )0

222

,1,1,

2

,,1,1 =

++

+ ++y

TTT

x



( ) ( )( ) ( ) ( ) ( ) ( ) 02 1,1,2

,1,1

2

,

22 =+++++ ++ jijijijiji TTxTTyTxy (4.74)

Then, the temperature at the internal nodes (nodes 5.6 and 7) can be calculated by

( ) ( ) ( ) ( )( ) ( )( )22

1,1,

2

,1,1

2

,2

0

xy

TTxTTyT

jijijiji

ji +

=+++= ++ (4.75)

For the bottom boundary nodes (j =1), since they are exposed to ambient temperature, the

general equation can be written as

( ) ( )0

222

1,0,2,

2

1,1,11,1 =

++

+ +y

TTT

x

TTT iiiiii(4.76)


29/48


From the convective boundary condition, we get

( )( )

fi

i,i,TTh

y

TTk =

1,20

2(4.77)

This leads to

( ) 2,1,02

ifii, TTTk

yhT +

= (4.78)


( ) ( )( ) ( ) ( ) ( )

( ) 022

22

2,2

1,11,1

2

1,

222

= ++

++

++ +

fi

iii

TkyhTx

TTyTk

yhxxy

(4.79)

or

( ) ( ) ( )

( ) ( )( ) ( )

++

+++

=+

k

yhxxy

Tk

yhTxTTy

Tfiii

i2

2

22

222

2,

2

1,11,1

2

1, (4.80)

Since the right surface is also exposed to ambient temperature, the equation for the right

boundary nodes (i = imax) can be expressed as

( ) ( )0

222

max,1max,1max,

2

max,,1max,1max =

++

+ ++y

TTT

x


where

( )fji

,jijiTTh

x

TTk =

+ max,

1max,1max

2

(4.82)

This leads to

( )jifji,ji TTT

k

xhT ,1maxmax,1max

2+ +

= (4.83)



30/48


( ) ( ) ( )( ) ( )

( ) ( ) 0

22

22

1max,1max,

2

,1max

2

max,

222

=++

+

+

++

+

jiji

jifji

TTx

TTk

xhyTxyy

k

xh

(4.84)

or

( ) ( ) ( )

( ) ( ) ( )( )

++

++

+

=+

222

1max,1max,

2

,1max

2

max,

22

22

xyyk

xh

TTxTTk

xhy

Tjijijif

ji (4.85)

Now, we demonstrate the treatment for the bottom right corner node (imax,1).

The discretized equation for this node is expressed as

( ) ( )0

222

1max,0max,2max,

2

1max,1,1max1,1max =

++

+ +y

TTT

x

TTT iiiiii(4.86)

Here, we need to replace two imaginary nodes (imax+1,1) and (imax, 0). Since the right

and bottom boundary are exposed to the ambient temperature, therefore

( ) 2max,1max,0max2

ifi,i TTTk

yhT +

= (4.87)

and

( ) 1,1max1max,11max2

+ +

= ifi,i TTTk

xhT (4.88)

Substitute Eq. (4.87) and (4.88) into Eq. (4.86) gives

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) 02222

222

2max,

2

1,1max

222

1max,

2222

=++

++

++

+

iif

i

TxTyTkyhx

kxhy

Txyk

yhx

k

xhy

(4.89)

Then


31/48


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )

++

+

++

+

=

2222

2max,

2

1,1max

222

1max,

222

2222

xyk

yhx

k

xhy

TxTyTk

yhx

k

xhy

Tiif

i (4.90)

Finally, since the problem is symmetric at the horizontal centerline of the fin, the

temperature at the top boundary nodes (node 1,2 and 3) can be set equal to the temperature

at the bottom boundary nodes.



width = 0.08;

height = 0.02;

h = 40;

k = 1;

t_air = 20;

t_left_wall = 300;

l=0; %

lmax=100;

imax=5;

jmax=3;

t(1:imax,1:jmax)=zeros;

imm1=imax-1;

jmm1=jmax-1;

dx = width/imm1;

dy = height/jmm1;

for j = 1:jmax


32/48


t(1,j) = t_left_wall;

end

for l=1:100

for i=2:imm1

j=2;

tnew=((t(i+1,j)+t(i-1,j))*dy^2 +...

(t(i,j+1)+t(i,j-1))*dx^2)/(2*(dy^2+dx^2));

t(i,j)=tnew;

end

for i=2:imm1

j = 1;

tnew=((t(i+1,j)+t(i-1,j))*dy^2 +

(2*t(i,j+1)+(2*dy*h*t_air)/k)*dx^2)/(2*(dy^2+dx^2)+((2*dy*h)/k)*dx^2);

t(i,j)=tnew;

end

for i=2:imm1

t(i,jmax)=t(i,1);

end

for j=2:jmm1

i = imax;

tnew=((t(i,j+1)+t(i,j-1))*dx^2 + (2*t(i-1,j)

+(2*dx*h*t_air)/k)*dy^2)/(2*(dy^2+dx^2)+((2*dx*h)/k)*dy^2);

t(i,j)=tnew;

end


33/48


j=1;

i=imax;

tnew=(2*(t(i-1,j)*dy^2+t(i,j+1)*dx^2) +...

(2*h*(dx*dy^2+dy*dx^2)/k)*t_air)/((2*h*(dx*dy^2+dy*dx^2)/k)

+2*(dy^2+dx^2));

t(i,j)=tnew;

t(imax,jmax)=t(imax,1);

end

4.3 Steady State Heat Conduction in Cylindrical Geometries.

Sometimes, it is desired to consider the governing equations expressed in the cylindrical

coordinate due to the nature of the problem. However, few modifications have to be made

for the finite different formulation in cylindrical geometries. To see this, let consider heat

conduction in a cylindrical geometry with the computational grids as shown below


34/48


Figure 4.10 Grid for radial conduction in a cylinder

Here, we consider one-dimensional heat conduction in radial direction and the governing

equations reduced to

01

2

2

=

+

r

T

rr

T(4.91)

Let the surface of the cylinder be maintained at a temperature of ST . For an axisymmetric

problem in cylindrical geometry, it can be shown that 0= rT at 0=r .

Now, consider a radial grid with constant size r . At a node i, the discretized form of the

heat conduction becomes

02

2 112

11 =

+

+ ++rr

TT

r

TTT

i

iiiii(4.92)

For the surface node (N+1), the prescribed temperature condition gives

SN TT =+1 (4.93)


35/48


The so called LHospital rule has to be applied at the center ofr= 0 to avoid 1/r

approaches as follow

2

2

1rT

rT

r = (4.94)

Therefore, the equation becomes

022

2

=r

T(4.95)

Therefore discretized equation at r = 0 is

0

2

2 2102

=

+r

TTT

(4.96)

Note that we consider an imaginary node at i = 0. This imaginary term can be replaced by

considering the boundary condition at r= 0

02

02 =r

TT(4.97)

which gives

20 TT = (4.98)

Then at the r= 0 boundary condition, we obtain

21 TT = (4.99)

Example 4.4

Consider a problem of heat conduction as shown in figure below. The inner side is heated

to the temperature of 30oC and outer side is heated to temperature of 150oC. Determine the

temperature distribution between these two boundaries.


36/48


Figure 4.11 Differentially heated 1D bar.

Solution

The discretized equation is given as

02

2 112

11 =

+

+ ++rr

TT

r

TTT

i

iiiii(4.100)

And the equation for the internal nodes is given by

( )

i

iiiii

r

TTr

TTT

42

1111 ++ ++

= (4.101)

Let r= 0.5m and mesh size is (M= 5), therefore

1.05

5.0==r (4.102)

The total number of grid points isM+1=6

The computer program code written in MATLAB is given as below


radius = 0.5;

mesh = 5;

deltaR = radius/mesh;

l = 0; %

lmax=100;

rmax=mesh+1;


37/48


t(1) = 30;

t(rmax) = 150;

for l =1:lmax

for r = 2:rmax-1

tnew(r) = (t(r+1)+t(r-1))/2 - deltaR*(t(r+1)-t(r-1))/(4*(r-

1)*deltaR);

t(r) = tnew(r);

end

end

The temperature distribution at steady state is obtained as in Tab. 4.3

Table 4.3 Temperature distribution for Example 4.4

Node 1 Node 2 Node 3 Node 4 Node 5 Node 6

30.0 34.8 49.2 73.2 106.8 150.0

Example 4.5

Consider a problem of heat conduction as shown in figure below. The inner side is

maintained at 50 oC and outer side is exposed to ambient temperature of 30oC and h =

50W/m2. Determine the temperature distribution between these two boundaries. (Consider

k= 30W/moC)


38/48


Figure 4.12 1D heated cylindrical bar exposed to ambient temperature.

Solution

Here, we consider one-dimensional heat conduction in radial direction and the governing

equations reduced to

01

2

2

=+

r

T

rr

T(4.103)

The discretized form of the governing equation is written as

02

2 11211 =+ +

++rrTT

rTTT

i

iiiii (4.104)

and the equation for the internal nodes is given by

( )

i

iiiii

r

TTr

TTT

42

1111 ++ +

= 0 (4.105)

For the internal surface node, the prescribed temperature condition gives

CT

0

1 500= (4.106)

Due to the convective heat transfer condition at the end point given as

( )fw TTh

r

Tk =

(4.107)

Discretise the above equation with central difference leads to

( )fi

ii TTk

h

r

TT=

+

max

1max1max

2(4.108)

or

( ) 1maxmax1max2

+ +

= ifii TTTk

rhT (4.109)

Substitute Eq. (4.109) into the discretized governing equation gives


39/48


022

22

1max

2

max

2

=+

+

+

+

+

iimacfimaci

imac

imac TrTk

hr

k

rhrT

kr

hr

k

rhr (4.110)

or

+

+

+

+

=

22

22

max

2

max

1maxmax

2

max

max

kr

hr

k

rhr

TrTk

hr

k

rhr

T

i

i

iifi

i (4.111)



radius = 0.5;

mesh = 5;

imax = mesh+1;

deltaR =radius/mesh;

t(1:imax) = zeros;

maxiter = 10000;

tf = 30;

h=50;

k=30;

t(1) = 50;

for iter = 1:maxiter

for i = 2:imax-1

tnew(i) = (t(i+1)+t(i-1))/2 + deltaR*(t(i+1)-t(i-1))/...

(4*(i-1)*deltaR);


40/48


end

i = imax;

ri=(i-1)*deltaR;

tnew(i)=((ri*2*deltaR*h+deltaR^2*h)*tf/k + 2*ri*t(i-1))/...

(ri*2*deltaR*h/k + deltaR^2*h/k +2*ri);

for i = 2:imax

t(i)=tnew(i);

end

end

4.3.1 Two-Dimensional Steady State Heat Conduction in Cylindrical Geometry.

The governing equation for 2D steady state heat conduction in cylindrical geometry is

given as

011

2

2

22

2

=

+

+

T

rr

T

rr

T(4.112)

Let imax andjmax be the number of nodes in the rand directions respectively. Using the

double notation, the discretized form of the governing equation at node (i,j) is given by

02

2

222

,1,1,,1,1

2

,,1,1 =

++

+

+ +++i

jijiji

i

jijijijiji

r

TTT

rr

TT

r

TTT(4.113)

Let the surface temperature be specified as

sji TT =max, (4.114)


41/48


for each angle j . At the centerr= 0, it is not possible to apply the LHospital rule. On

this mesh, the heat conduction in Cartesian coordinates which is discretized as

0222

02

0 = ++ + rTTT

rTTT SNWE (4.115)

is used. Here we consider that ryx ==

Problem 4.6

The temperature at the surface of half cylinder is fixed at 100oC while the lower surface is

maintained at 300oC. Compute the temperature for nodes shown in Figure 4.13

Figure 4.13 Differentially heated surface of half cylinder

Solution

The governing equation for this case can be written as

0

112

2

22

2

=

+

+

T

rr

T

rr

T

(4.116)

Applying second order central difference and Eq. (4.116) can be discretized into the

following algebraic expression as follow


42/48


02

2

2

22

,1,1,,1,1

2

,,1,1 =

++

+

+ +++i

jijiji

i

jijijijiji

r

TTT

rr

TT

r

TTT(4.117)

For the internal nodes, Eq. (4.117) can be rewritten as

1,222

2

1,222

2

,1222

222

,1222

222

,

44

2

44

2

44

2

44

2

+

+

+

++

++

+++

=

ji

i

ji

i

ji

i

iiji

i

iiji

Trr

rT

rr

r

Trr

rrrT

rr

rrrT

(4.118)

or

1,1,,1,1, ++ +++= jijijijiji CTCTBTATT (4.119)

where

222

2

222

222

222

222

44

2

44

2

44

2

rr

rC

rr

rrrB

rr

rrrA

i

i

ii

i

ii

+

=

+

=

++

=

(4.120)



radius = 0.05;

theta = pi;

k = 20;

t_wall1 = 100;

t_wall2 = 300;

l=0;

lmax=10000;

imax=3;


43/48


jmax=7;

imm1=imax-1;

jmm1=jmax-1;

dr = radius/imm1;

dthe = theta/jmm1;

t(imax,jmax)=zeros;

%boundary condition

for j = 1:jmax

t(imax,j) = t_wall1;

end

for i = 1:imm1

t(i,1) = t_wall2;

t(i,jmax) = t_wall2;

end

t0 = 300;

for l=1:lmax

for i=1:imm1

for j = 2:jmm1

A = (2*((i*dr)^2)*(dthe^2)+(i*dr)*dr*(dthe^2))/(4*((i*dr)^2)*dthe^2

+4*dr^2);

B = (2*((i*dr)^2)*(dthe^2)-(i*dr)*dr*(dthe^2))/(4*((i*dr)^2)*dthe^2

+4*dr^2);

C = 2*(dr^2)/(4*((i*dr)^2)*dthe^2 + 4*dr^2);


44/48


if i == 1

tnew(i,j)=A*t(i+1,j) + B*t0 +C*t(i,j+1)+C*t(i,j-1);

else

tnew(i,j)=A*t(i+1,j) + B*t(i-1,j) +C*t(i,j+1)+C*t(i,j-1);

end

t(i,j)=tnew(i,j);

end

end

end

disp(t)

Problem 4.7

The half cylinder has ./3 mWk= oC and is exposed to the convection environment at 20oC.

The lower surface is maintained at 100oC. Compute the temperature for nodes shown in

Figure 4.14

Figure 4.14 Heated half cylinder and exposed to ambient temperature


45/48


Solution

The governing equation for this case can be written as

011

2

2

22

2

=++ T

rr

T

rr

T(4.121)

Applying second order central difference, Eq. (4.121) can be discretized into the following

algebraic expression as follow

02

2

222

,1,1,,1,1

2

,,1,1 =

++

+

+ +++i

jijiji

i

jijijijiji

r

TTT

rr

TT

r

TTT(4.122)

For the internal nodes, Eq. (4.122) can be rewritten as

1,1,,1,1, ++ +++= jijijijiji CTCTBTATT (4.123)

where A, B and C are expressed in Eq. (4.120)

Fig. 4.15 Coordinates of node and the center and its neighbor

At the center node of the surface, we have to consider a Cartesian based governing

equation which is

02

2

2

2

=

+

y

T

x

T(4.124)

After discretization, Eq. (4.124) becomes (See Fig. 4.15)

( )0

222

021max,1

2

0max,11,1 =

++

+ +r

TTT

r

TTT jNj(4.125)

Here ryx ==


46/48


Then

( )

4

21max,1max,11,1

0

++++= jNjTTTT

T (4.126)

Since the boundary is exposed to the ambient temperature where

( )fTTh

y

Tk =

0 (4.127)

or

( ) ( )f

jNTT

k

h

y

TT=

+0

21max,1

2(4.128)

and therefore

( ) ( )fjN TTk

yhTT

= + 021max,1

2(4.129)

Substitute into Eq. (4.128) and rearrange gives

( )

+

+++

=+

k

yh

Tk

yhTTT

Tfjj

214

22 21max,1max,11,1

0 (4.130)

Here, ry = then

( )

+

+++

=+

k

rh

Tk

rhTTT

Tfjj

214

22 21max,1max,11,1

0 (4.131)

The discretized equation for the top boundary nodes can be written as

0,2,1,11,11, iiiii CTCTBTATT +++= + (4.132)

Since these boundary nodes are exposed to the ambient temperature where


47/48


( )fw TT

k

h

r

T=

(4.133)

or in discretized form, Eq. (4.133) becomes

( )fi

i

iiTT

k

h

r

TT=

1,

2,0,

2 (4.134)

and therefore

( )fiiii TT

k

hrTT = 1,2,0, 2 (4.135)

Substitute Eq. (4.135) into Eq. (4.134) yields

+++

+

= + fiiiii

i Tk

hCrCTBTAT

hCrk

kT

22

22,1,11,11, (4.136)


clear all;clc;close all;radius = 0.05;theta = pi;k = 3;t_wall1 = 100;

tf = 20;h=50;imax=3;jmax=7;imm1=imax-1;jmm1=jmax-1;dr = radius/imm1;dthe = theta/jmm1;t(imax,jmax)=zeros;%boundary condition

for j = 1:jmaxt(imax,j) = t_wall1;endfor l=1:100000t0 = (t(1,1)+ t(1,jmax)+2*t(1,(jmax+1)/2)+2*dr*h*tf/k)/(4+2*dr*h/k); for i=1:imm1


48/48


for j = 1:jmaxA = (2*((i*dr)^2)*(dthe^2)+(i*dr)*dr*(dthe^2))/

(4*((i*dr)^2)*dthe^2 +4*dr^2);B = (2*((i*dr)^2)*(dthe^2)-(i*dr)*dr*(dthe^2))/

(4*((i*dr)^2)*dthe^2 +4*dr^2);

C = 2*(dr^2)/(4*((i*dr)^2)*dthe^2 + 4*dr^2);

if i == 1 && j == 1tnew(i,j)=(A*t(i+1,j) + B*t0

+2*C*t(i,j+1)+2*C*(i*dr)*dthe*h*tf/k)/...(1+2*C*(i*dr)*dthe*h/k);

elseif i ==1 && j~=1 && j~=jmaxtnew(i,j)=A*t(i+1,j) + B*t0 +C*t(i,j+1)+C*t(i,j-1);

elseif j ==1 && i ~=1tnew(i,j)=(A*t(i+1,j) + B*t(i-1,j)

+2*C*t(i,j+1)+2*C*(i*dr)*dthe*h*tf/k)/...

(1+2*C*(i*dr)*dthe*h/k); elseif i ==1 && j ==jmax

tnew(i,j)=tnew(1,1); elseif i ~=1 && j == jmax

tnew(i,j)=tnew(2,1); else

tnew(i,j)=A*t(i+1,j) + B*t(i-1,j)+C*t(i,j+1)+C*t(i,j-1); end

t(i,j)=tnew(i,j); end

endenddisp(t')

Date post:	05-Apr-2018
Category:	Documents
Upload:	rozanti-a-hamid
View:	213 times
Download:	0 times

Comp Heat Transfer4 Repaired)

Documents