# Problem1.4.1g - Problem 1.4.1(g The differential equation...

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Problem 1.4.1(g) The differential equation and BCs at steady state are given by (1) 2 u ÅÅÅÅÅÅÅÅÅ x 2 = 0, u H 0 L = T, u ÅÅÅÅÅÅÅ x H L L + u H l L = 0 The general solution (after integrating twice) is (2) u H x L = C 1 x + C 2 Applying the BC at x=0 gives (3) C 2 = T, At x=L we have

Unformatted text preview: (4) C 1 + C 1 L + T = Solving for C 1 gives (5) C 1 = -T ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H 1 + L L Thus the steady state solution is (6) u H x L = -T ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H 1 + L L x + T...
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