EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 29, Problem 7P
To determine

To calculate: The temperature on the heated plate with fixed temperature at the boundaries and round corners at upper right and lower left corners; considering the value of weighing factor for overrelaxation as 1.5 and performing iteration up to εs=1%.

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Answer to Problem 7P

Solution:

Heated plate having rounded upper right and lower left corners has the temperature distribution with the desired level of accuracy as,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 7P

Explanation of Solution

Given Information:

The fixed temperature at top and right boundary is 100 °C and the fixed temperature at the left and lower boundary is 50 °C.

For overrelaxation the weighing factor is 1.5.

Limiting condition for iteration is εs1%

Formula used:

For a square grid, that is, Δx=Δy, the Laplacian difference equation is written using Liebmann’s method as,

Ti,j=Ti+1,j+Ti1,j+Ti,j+1+Ti,j14

With i varying from 1 to m and j varying from 1 to n.

Overrelaxation is applied using the following expression after each iteration,

Ti,jnew=λTi,jnew+(1λ)Ti,jold

In the above expression, Ti,jold and Ti,jnew being the Ti,j values from the previous and present iteration.

The percent relative error is calculated using the following expression,

|(εa)i,j|=(|Ti,jnewTi,joldTi,jnew|×100)%

Laplacian equation for a node adjacent to an irregular boundary is solved for temperature distribution as,

2Δx2[Ti1,jTi,jα1(α1+α2)+Ti+1,jTi,jα2(α1+α2)]+2Δy2[Ti,j1Ti,jβ1(β1+β2)+Ti,j+1Ti,jβ2(β1+β2)]=0

Calculation:

Consider, Δx=Δy.

Consider the value of the parameters as,

α1=0.732α2=1β1=0.732β2=1

The temperature at the node (4,2) adjacent to the lower left round corner or the irregular boundary is calculated as,

2Δx2[T3,2T4,20.732(0.732+1)+T5,2T4,21(0.732+1)]+2Δy2[T4,1T4,20.732(0.732+1)+T4,3T4,21(0.732+1)]=0[T3,2T4,20.732(0.732+1)+T5,2T4,21(0.732+1)]+[T4,1T4,20.732(0.732+1)+T4,3T4,21(0.732+1)]=0

Thus,

4T4,2+0.8453T5,2+0.8453T4,3=1.1547(T3,2+T4,1)

Further, the temperature at the node (2,4) adjacent to the upper right round corner or the irregular boundary is calculated as,

2Δx2[T1,4T2,40.732(0.732+1)+T3,4T2,41(0.732+1)]+2Δy2[T2,3T2,40.732(0.732+1)+T2,5T2,41(0.732+1)]=0[T1,4T2,40.732(0.732+1)+T3,4T2,41(0.732+1)]+[T2,3T2,40.732(0.732+1)+T2,5T2,41(0.732+1)]=0

Thus,

4T2,4+0.8453T3,4+0.8453T2,5=1.1547(T1,4+T2,3)

Liebmann’s method is used to solve the temperature on the heated plate iteratively in MATLAB, until the percent relative error falls below 1%.

Use the following MATLAB code to implement the Liebmann’s method and the equation at the nodes (4,2)&(2,4) to solve for the temperature distribution on the heated plate with fixed boundary conditions as,

clear;clc;

% Enter the initial temperature distribution

T = zeros(5,5);

% Enter the initial conditions

T(1,1:4) = 100;

T(2:5,5) = 100;

T(5,2:4) = 50;

T(1:4,1) = 50;

ans

n = length(T(:,1));

m = length(T(1,:));

% Enter the Weighing factor

lambda = 1.5;

iteration = 0;

A = ((1/(0.732*1.732))*2/((1/(0.732*1.732))+(1/(1.732))));

B = ((1/(1.732))*2/((1/(0.732*1.732))+(1/(1.732))));

% Start the loop and run it until Ea < 1

while 1

for i = 2:m-1

for j = 2:n-1

Told(i,j) = T(i,j);

% for lower left and upper right corner nodes

if (i == 4 && j == 2) || (i == 2 && j == 4)

T(i,j) = (-(A*(T(i-1,j)+T(i,j-1)))-(B*T(i+1,j))...

-(B*T(i,j+1)))/(-4);

else

% Liebmann's method for rest of the interior nodes

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

% Percent relative error

Ea(i,j) = abs((T(i,j)-Told(i,j))/T(i,j))*100;

T(i,j) = (lambda*T(i,j))+((1-lambda)*Told(i,j));

end

end

end

iteration = iteration+1;

if max(Ea(:,:)) < 1

break

end

end

T

iteration

Execute the above code to obtain the temperature distribution and the number of iterations performed, in the output as shown below,

T =

50.0000 100.0000 100.0000 100.0000 0

50.0000 75.0538 85.8151 92.8540 100.0000

50.0000 64.3565 75.0200 85.7216 100.0000

50.0000 57.1402 64.2716 75.0977 100.0000

0 50.0000 50.0000 50.0000 100.0000

iteration =

9

Hence, the temperature on the heated plate with desired accuracy is achieved after 9 iterations as obtained in the above MATLAB output.

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