This document contains the analysis of two 1-D steady heat conduction problems using the Finite Difference Method (FDM). The first case considers a metal rod with a base temperature of 100°C and a tip temperature of 30°C. The second case considers the same rod but with the tip insulated, so the temperature gradient is calculated to the second last node. Both cases are solved for varying nodal discretizations and the temperature profiles along the rod length are plotted.

1. www.engineeringwithsandeep.com| Student Assignment
Assignment No. 07
1-D Steady Heat conduction FDM
14.04.2021
Shivam Choubey
Student No: CFDB30321004

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Case1:
T base =100 C and T tip = 30 C
%% 1-D Steady Heat conduction FDM Code
% 1-D steady heat conduction problem
% metal rod which is 20 cm long
%Case1: T base =100 C and T tip = 30 C
clc
clear all;
L=20/100;% length convert into meter
N=80;%N=10 ;N=20;N=40;N=50;N=80;
dx=linspace(0,L,N);
Tb=100;
Tend=30;
T=zeros(N,1);%Initialization
T(1)=Tb;% base temp
T(N)=Tend;%Tip temp

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k=100;% Correction loop
for j=1:k
for i=2:N-1
T(i)=(T(i+1)+T(i-1))/2;
end
T(N)=Tend;%Tip temp
end
plot(dx,T);
hold on
xlabel("Length (m)")
ylabel("Temp")
title("Length Vs Temp")
legend("N=10","N=20","N=40","N=50","N=80")
hold on

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Case 2:
T base =100C and Tip is insulated
%% 1-D Steady Heat conduction FDM Code
% 1-D steady heat conduction problem
% metal rod which is 20 cm long
%Case1: T base =100 C and T tip = 0 C
clc
clear all;
L=20/100;% length convert into meter
N=100;%N=40 N=50 N=80 N=100
dx=linspace(0,L,N);
Tb=100;
Tend=0;
T=zeros(N,1);%Initialization
T(1)=Tb;% base temp
T(N)=Tend;%Tip temp
k=100;% Correction loop
for j=1:k

5. www.engineeringwithsandeep.com| Student Assignment
for i=2:N-1
T(i)=(T(i+1)+T(i-1))/2;
end
%T(N)=Tend;%Tip temp
T(N)=T(N-1);%condition for insulated tip
end
plot(dx,T);
hold on
xlabel("Length (m)")
ylabel("Temp")
title("Length Vs Temp")
legend("N=10","N=20","N=40","N=50","N=80","N=100")
hold on