Case 2:Temperature Distrbution In Thin Rod The differential equation derives from a heat balance for along thin rod as follow: d?T + h' (Ta – T) = 0.. .... eq(1) - dx2 Where T=temperature ("C), =distance along the rod (m), h'=a heat transfer coefficent between the rod and the ambient air (m2), and T=the temperature of the surrounding air (°C).This equation can be transformed into a set of linear algebraic equations by using a finite dividd difference approximation for the second derivative as follow: d?T Ti+1 - 27; + Ti-1 .. .eq(2) ......... dx? Δx2 Where Ti designates the temperature at node i. This approximation can be substituded into eq(1) to give: -Ti-1+ (2+ h'Ax?)Ti – Ti+1 = h'Ax2Ta. eq(3) This equation can be written for each of the interior nodes of the rod resulting in a tridiagonal system of equations. The first and the last nodes at the rods ends are fixed by boundary conditions. T(r=0)=200 °C, T(x=10)= 40 °C, h'=0.02 m?, T=10 °C, Ar=2m. If this system of linear equation can be solved using Gauss-Seidel and Jacobi iterations methods. Please answer the following: 1. Develop a MATLAB program to solve the system above using both Gauss-Seidel and Jacobi iteration methods after n of iterations. 2. Plot the values of each tempratures (T1,T2,T3,T4) versus iteration number in both solutions. Compare the values you get from both solution methods. 3. Plot the values of the relative approximate error of each temperatures (Ea) versus iteration number in both solutions. Compare the values you get from both solution methods. 4. Use LU decomposition to find the exact solution {Ti} and compare your exact values to those you got from point 1. 5. Plot the temperatures you got from point 4 along the rod.
Case 2:Temperature Distrbution In Thin Rod The differential equation derives from a heat balance for along thin rod as follow: d?T + h' (Ta – T) = 0.. .... eq(1) - dx2 Where T=temperature ("C), =distance along the rod (m), h'=a heat transfer coefficent between the rod and the ambient air (m2), and T=the temperature of the surrounding air (°C).This equation can be transformed into a set of linear algebraic equations by using a finite dividd difference approximation for the second derivative as follow: d?T Ti+1 - 27; + Ti-1 .. .eq(2) ......... dx? Δx2 Where Ti designates the temperature at node i. This approximation can be substituded into eq(1) to give: -Ti-1+ (2+ h'Ax?)Ti – Ti+1 = h'Ax2Ta. eq(3) This equation can be written for each of the interior nodes of the rod resulting in a tridiagonal system of equations. The first and the last nodes at the rods ends are fixed by boundary conditions. T(r=0)=200 °C, T(x=10)= 40 °C, h'=0.02 m?, T=10 °C, Ar=2m. If this system of linear equation can be solved using Gauss-Seidel and Jacobi iterations methods. Please answer the following: 1. Develop a MATLAB program to solve the system above using both Gauss-Seidel and Jacobi iteration methods after n of iterations. 2. Plot the values of each tempratures (T1,T2,T3,T4) versus iteration number in both solutions. Compare the values you get from both solution methods. 3. Plot the values of the relative approximate error of each temperatures (Ea) versus iteration number in both solutions. Compare the values you get from both solution methods. 4. Use LU decomposition to find the exact solution {Ti} and compare your exact values to those you got from point 1. 5. Plot the temperatures you got from point 4 along the rod.
C++ for Engineers and Scientists
4th Edition
ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter6: Modularity Using Functions
Section6.2: Returning A Single Value
Problem 8E
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