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Heat Transfer: Finite Difference method using MATLAB
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[QUOTE="aznkid310, post: 2175245, member: 106903"] [h2]Homework Statement [/h2] I have uploaded the problem statement and corresponding programs. Reader is the problem statement Reader1 is another file that is suppose to help [h2]Homework Equations[/h2] I'm just having trouble getting started. I can do the analytical part on paper, but I don't know how to implement this in MATLAB. In the program for problem 1: What is the 'A' and 'C' matrix? I can get some values on paper, but, for example, what is A(i, i+1)? I realize this this the temperature to the left of the node I am interested in, but do I type in some sort of equation? Similarly for A(i,i), A(i, i-1), C(n)... Is C(1) = -200 from my solution? It seems I have to do what is says in the reader1 file, but I don't know how to use it [h2]The Attempt at a Solution[/h2] For #1a: T(x) = C1x + C2 Boundary Conditions: T(0) = 500 T(L = 1m) = 300 Thus T(x) = 500 - 200x q = -kA(dT/dx) = 37200 Watts, where k = 186 W/m.k energy balance for [B]internal nodes[/B]: q1 + q2 +qV = 0 q1 = kA[ (T_i-1) - T_i] q2 = kA[(T_i+1) - T_i] qV = dx = internal heat generation assuming dy=dz = infinite Finite Diff. eqn: (T_i-1) - (2T_i) + (T_i+1) = 0 For the left [B]boundary node[/B]: q_left side = 0 b/c T = 500k is constant? therefore q_right side + qV = 0 q_right side = kA[(T_i+1) - T_i] and qV is the same from before [B]Right Boundary Node[/B]: q_left side = kA[(T_i-1) - T_i] and qV stays the same How is this implemented in MATLAB? For #1b: Boundary Conditions: T(0) = 500 -kA(dT/dx) @ x=1m = hA(Ts -T_inf) [Convective BC] Assuming C1 is the same from part a (is this correct?): T_inf = 462.8 K [/QUOTE]
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Heat Transfer: Finite Difference method using MATLAB
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