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 1P

Use Liebmann's method to solve for the temperature of the square heated plate in Fig. 29.4, but with the upper boundary condition increased to 150 ° C and the left boundary insulated. Use a relaxation factor of 1.2 and iterate to ε S = 1 % .

Expert Solution & Answer
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To determine

To calculate: The temperature of the square heated plate in the provided figure by Liebmann’s method with the upper boundary condition increased to 150°C and the left boundary insulated;with the use of a relaxation factor of 1.2 and with iteration to εs=1%.

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 1P , additional homework tip  1

Answer to Problem 1P

Solution:

The temperature of the square heated plate is,

0°C67.8609  88.3949  85.715533.0467  50.0069  54.465914.2989  24.1186  32.1467 0°C150°C50°C

Explanation of Solution

Given information:

The upper boundary condition increased to 150°C and the left boundary insulated and with the use of a relaxation factor of 1.2 and with iteration to εs=1%. The provided figure is,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 1P , additional homework tip  2

Calculation:

The following formula is used to calculate the temperature at (i,j) node as,

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

And the percent relative errors is,

|(εa)i,j|=|Ti,jnewTi,joldTi,jnew|100%

With the help of the above formula, the following code implements the procedure of Liebmann’s method to solve for the temperatures on the platewith the given conditions, that is, theleft boundary is insulated and the upper boundary raised to 150°C(from the previous 100 °C). The code can be written with the following script in MATLAB:

Code:

clear;

% initial boundary conditions are given

T =zeros(5);

T(:,1)=0;% left boundary condition

T(5,:)=150;% upper boundary condition

T(:,5)=50;% right boundary condition

T(1,:)=0;% bottom boundary condition

% relaxation and error threshold are defined

lambda =1.2;

epsilon =0.01;

% error matrix definition is given.

errors =ones(3);

iter=0;

whilemax(errors(:))> epsilon

for j =2:4

fori=2:4

Told = T(i,j);

% compute new T at (i, j)

T(i,j)=0.25*(T(i+1,j)+T(i-1,j)+T(i,j+1)+T(i,j-1));

% apply overrelaxation

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

T(i,j)=Tnew;

% compute error

errors(i-1,j-1)= abs(Tnew- Told)/abs(Tnew);

end

end

% print result values with iteration

iter=iter+1;

T(2:4,2:4)

maximun_ea=max(errors(:))*100;

disp(maximun_ea)

% print maximum error

end

Output:

On executing the above script, the program finds the stopping condition on the seventh iteration with a maximum error of 0.5169%.

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 1P , additional homework tip  3

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 1P , additional homework tip  4

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 29, Problem 1P , additional homework tip  5

From the above output, the final result can be shown as:

0°C67.8609  88.3949  85.715533.0467  50.0069  54.465914.2989  24.1186  32.1467 0°C150°C50°C

And the maximum Error is 0.5169.

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