 # Steady State Heat Conduction

## Transcribed Text

Application to Steady-State Heat Conduction. Use function for LU decomposition and substitution in order to solve a problem involving 1-D steady-state heat conduction. Consider a slender bar of length L. On one end (x = 0) it is subjected to a fixed temperature, To, while at the other end (. = L) it is insulated. There is an external heat source, with heat per unit length, dc(I). The governing equation for the temperature distribution, T(x), is d²T dT T(0) = To, (L) = O, (4) da where 6 is the thermal conductivity. You should obtain a linear set of equations, AT = b, for the temperatures at the nodes. Your routines must set up the problem-including setting up the matrix A for the discrete set of equations-and solve it and plot results. In your study, set L - 1 m and To = 500 K. The thermal conductivity is N = 40W/(m K). Discretize the problem on a grid with 200 points. The heat source has the form (5) Use 40 = 1000W/m². Try two different values for 20 (where both values are between 0 and L) and two different values of OT (I suggest a = 0.5 and a = 0.05). Remember that the point of using LU decomposition is that this only needs to be performed once for a given matrix A. Only the forward and backward substitution need to be repeated when the right-hand side changes.

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%Application to steady state heat conduction

%Clearing the screen, workspace and closing all plots
clear
clc
close all
%%

%Defining parametres
L=1; %L=1m
T0=500; %T0=500K
k=40; %k=40W/(m*K)
points=200;
q0=1000; %q0=1000W/m^2
sigma1=0.5; %sigma=0.5
sigma2=0.05; %sigma=0.05
x01=0.3; %x=0.3m
x02=0.8; %x=0.8m
n=points;
%%...
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