Chemistry, 16.10.2019 21:20 tiffanybrandy23
Consider heat conduction through the wall of a pipe with inside wall temperature of 200oc and outside temperature of 80oc. the wall is 0.05 m thick with inside radius r0 of 0.05 m and outer radius rn of 0.1 m. thermal conductivity of the wall is 1w/moc. the differential equation which describes the radial temperature distribution is dt dr + r d 2t dr 2 = 0 solving the above equation with finite difference method on n equally-spaced node points along the radial direction, together with application of temperature boundary conditions at inner and outer wall yields the following system of equations: t1 = 200 ri βr 2 β 1 βr tiβ1 β 2ri βr 2 ti + ri βr 2 + 1 βr ti+1 = 0 i = 2, 3, n β 1 tn = 80 ri is radial position of the nodes and can be written as ri = r0 + βr (i β 1) i = 1, 2, n radial spacing between the nodes βr can be computed as: βr = rn β r0 n β 1 (a) solve the temperature values t using thomas method with number of nodes n = 11. remember that radius ri is not constant. provide the m-files. (b) plot the temperature profiles, i. e. t versus ri . make sure the axis are properly labeled.
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Consider heat conduction through the wall of a pipe with inside wall temperature of 200oc and outsid...
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