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//Eg-14.3
//pg-583
clear
clc
close()
//Analytically solving the given equation using central difference formula we get 3 equations at 3 internal points
// At point 1 T0 - 2*T1 + T2 = -3.125 (1)
// At point 2 T1 - 2*T2 + T3 = -3.125 (2)
// At point 3 T2 - 2*T3 + T4 = -3.125 (3)
// Using BC 1 T-1 = T1
// Using BC 2 T5 = T3 - 0.25*T4 + 75;
// using BC 1 in (1), we get T1 - T0 = -1.5625 (4)
// using the value of T5 in the BC gives 2*T3 - 2.25*T4 = -78.125 (5)
//Solving these equations gives the values of T at different points
A = [1 -2 1 0 0;0 1 -2 1 0;0 0 1 -2 1;1 -1 0 0 0;0 0 0 2 -2.25];
B = [-3.125;-3.125;-3.125;1.5625;-78.125];
X = inv(A)*B;
x = 0:0.25:1;
plot(x,X,'ks')
//Analytically
L = 1;
x1 = 0:0.01:1;
T = 300 + 50*L^2/(2)*(1-(x1/L)^2 + 2*2);
plot(x1,T)
xlabel('x(m)')
ylabel('T(K)')
legend('FD solution','Analytical Solution')
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