X x e e t analytical solution heat transfer

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xxeeTAnalytical Solution:(Heat transfer coefficient(
79The Shooting MethodConverts the boundary value problem to initial-value problem. A trial-and-error approach is then implemented to solve the initial value approach.For example, the 2ndorder equation can be expressed as two first order ODEs:An initial value is guessed, say z(0)=10.The solution is then obtained by integrating the two 1storder ODEs simultaneously.)(aTThdxdzzdxdT
80Using a the RK method. )]43([';)(')43,43(;),(;32314343;)43,43(;),(;32311,2,1,1,2,1,2,1,11,2,1,1,2,1,2,1,1hkTThkTThkorhkThxfkTxfkhkkzzhzzhkzkzkorhkzhxfkzxfkhkkTTziaziazTiizziizzzziiiiTiTiTTiiTTiiTTTTii)(aTThdxdzzdxdThkakazzhkakaTThaveWezziiTTii)()(:2,21,112,21,11
81Using Goal seek - one obtains T(10)=200. for z(0)=18.11654Using a 4thorder RK method with a step size of 2:T(10)=168.3797.This differs from T(10)=200. Therefore a new guess is made, z(0)=20and the computation is performed again.z(0)=20T(10)=285.8980Since the two sets of points, (z, T)1and (z, T)2,are linearly related, a linear interpolation formula is used to compute the value of z(0)as 12.6907to determine the correct solution.
Using a linear interpolation formula between the valuesof z(0), determine a new value of z(0(Recall:first estimate z(0) = 10 T(20) = 168.38second estimate z(0)=20 T(20) = 285.90What is z(0) that would give us T(20)=200?1502002503000510152025z(0(T(20(
z010201028590168 38200168 3812 69....1502002503000510152025z(0(T(20(
z010201028590168 38200168 3812 69....1502002503000510152025z(0(T(20(d TdxhTTdTdxzdzdxhTTaa220''We can now use this to solve the first order ODE
85Graphical illustration of the shooting method
Finite Difference MethodsThe finite divided difference approximation forthe 2nd derivative can be substituted into thegoverning equation.d TdxTTTxd Tdxh TTTTTxh TTiiiaiiiai22112221122020''
TTTxh TTThxTThx Tiiiaiiiia1121212202'''Collect termsWe can now apply this equation to each interior nodeon the rod.Divide the rod into a grid, and consider a “node” to beat each division. i.e.. x = 2mT1T2L = 10 mx = 2 mConsider:L = 10 mTa= 20T(0) = 40T(10) = 200h’ = 0.01
aiiiTxhTTxhT2121''2T(0(T(10(L = 10 mx = 2 m
aiiiTxhTTxhT2121''2T(0(T(10(x=0 2 4 6 8 10i=0 1 2 3 4 5Notice the labeling for numbering xand i
aiiiTxhTTxhT2121''2T(0(T(10(x=0 2 4 6 8 10i=0 1 2 3 4 540200Note also that the dependent values are known at the boundaries (hence the term boundary value problem(
aiiiTxhTTxhT2121''2T(0(T(10(x=0 2 4 6 8 10i=0 1 2 3 4 540200Apply the governing equation at node 18.4004.28.004.240''2212122120TTTTTxhTTxhTa
aiiiTxhTTxhT2121''2

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