Numerical Methods For Engineers, 7 Ed
Numerical Methods For Engineers, 7 Ed
7th Edition
ISBN: 9789352602131
Author: Canale Chapra
Publisher: MCGRAW-HILL HIGHER EDUCATION
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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,

Numerical Methods For Engineers, 7 Ed, 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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