Chemistry
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 βˆ†r 2 βˆ’ 1 βˆ†r tiβˆ’1 βˆ’ 2ri βˆ†r 2 ti + ri βˆ†r 2 + 1 βˆ†r ti+1 = 0 i = 2, 3, n βˆ’ 1 tn = 80 ri is radial position of the nodes and can be written as ri = r0 + βˆ†r (i βˆ’ 1) i = 1, 2, n radial spacing between the nodes βˆ†r can be computed as: βˆ†r = rn βˆ’ r0 n βˆ’ 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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