Question 46 - In a steady state one dimensional conduction with no heat generation, the differential equation is d / dx (k dT / dx) = 0. Prove that T(x) = ax + b, where k, a and b are constants. (b) At x = 0, T = c and at x = L, T = d. Prove that T(x) = (d - c) x / L + c for boundary conditions.
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Answer Posted / kang chuen tat (malaysia - pen
Answer 46 - When d / dx (k dT / dx) = 0, d (dT) / [ (dx) (dx) ] = 0. Integrate both sides gives dT / dx = a. Second integration gives T(x) = ax + b for both sides (proven). (b) T(0) = a(0) + b = b = c. T(L) = d = aL + c then a = (d - c) / L. Substitute in T(x) = ax + b gives T(x) = (d - c) x / L + c (proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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