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### proj1s99sol

Course: ME 303, Fall 2009
School: W. Alabama
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OF PROJ1S99SOL.TEX UNIVERSITY WATERLOO Department of Mechanical Engineering ME 303 Advanced Engineering Mathematics M.M. Yovanovich Project 1 Solution, June 4, 1999 Given the linear, second order nonhomogeneous PDE: ! 1 @ r @T + S = 1 @T ; t 0; 0 r a r dr @r k @t with parameters: S W=m3 , and k W=m K . The units of the dependent variable is T K , The units for the other variables are: r m ; t s , and a m ....

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OF PROJ1S99SOL.TEX UNIVERSITY WATERLOO Department of Mechanical Engineering ME 303 Advanced Engineering Mathematics M.M. Yovanovich Project 1 Solution, June 4, 1999 Given the linear, second order nonhomogeneous PDE: ! 1 @ r @T + S = 1 @T ; t 0; 0 r a r dr @r k @t with parameters: S W=m3 , and k W=m K . The units of the dependent variable is T K , The units for the other variables are: r m ; t s , and a m . Boundary conditions and initial condition are @T 0; t = 0; T a; t = 0 and T r; 0 = 0; 0 r a @r There is no convenient temperature scale and time scale in the problem statement. 1 What are the units of ? The units of the three terms are identical. The units can be obtained from the rst or second term. The units of the second term source term of the PDE are W=m3 = W=m K = K=m2 . Since the units of the transient term energy storage term are identical we have 1K = K s m2 The units of the thermophysical parameter are m2=s . 2 Obtain nondimensional form of PDE, BCs and IC. Use the following dimensionless parameters: r; r ; = T T t ; = tt ; =a r r 1 where Tr is an arbitrary reference temperature, tr is an arbitrary reference time, and a is the radius of the solid circular cylinder. Nondimensionalize the rst term. Set T = Tr where Tr is a constant. @T = @ T @ = Tr @ r @r @ @r a @ Now, r @T = a Tr @ = Tr @ @r a@ @ Therefore, Finally, @ r @T = @ @r @r @ ! ! Tr @ @ ! @ = Tr @ @r a @ ! @ @ ! 1 @ r @T = 1 Tr @ r @r @r a a@ @ @ r = T2 1 @ a @ @ @ ! Note which parameters determine the units at each step of the nondimensionalization process. Nondimensionalize the transient term in a similar manner. @T = @ Tr @ = Tr @ @t @ @t tr @ The units are determine by the parameters: Tr =tr . The PDE can now be written as ! Tr 1 @ @ + S = 1 Tr @ a2 @ @ k tr @ The terms are still dimensional. Observe that the units of the rst term are determined by the parameters Tr =a2, and the units of the third term are determined by the parameters Tr=tr . Multiple through by a2=Tr to get the nondimensional form of the PDE ! 1 @ @ + Sa2 = a2 @ ; 0; 0 1 @ @ kTr tr @ Note that by inspection of the second and third terms of the PDE, we that nd the units of the group S a2=k are K , and the units of the group a2= are s . 2 Tberefore S a2=k represents a temperature scale, and a2= represents a time scale of the system. The nondimensional form of the BCs are @T 0; t = @ Tr 0; @ = Tr @ 0; = 0 or @ 0; = 0 @r @ @r a @ @ and T a; t = Tr 1; = 0 or 1; = 0 The nondimensional form of the IC is T r; 0 = Tr ; 0 = 0; 0 r a or ; 0 = 0; 0 1 3 Set the reference temperature: Tr = Sa2=k and the reference time: tr = a2= in the PDE to obtain the nal dimensionless form without parameters: ! 1 @ @ +1= @ ; 0; 0 1 @ @ @ with homogeneous BCs and IC: @ 0; = 0; 1; = 0 and ; 0 = 0; 0 1 @ 4 Obtain the solution of the steady-state case where @ =@ = 0. The PDE becomes the ODE with ! 1 @ @ + 1 = 0; 0 1 @ @ and homogeneous BCs: @ 0; = 0; 1; = 0 @ This ODE can be integrated twice in a straight forward manner. Here are the steps: ! @ @ =, @ @ Integrate once to get @ =, 2 +C 1 @ 2 3 Dividing by gives the derivative: @ = , + C1 @ 2 The rst BC at = 0 can be used now to eliminate the second term which is unbounded at = 0. Set C1 = 0 to give @ =, @ 2 Integrate a second time to get Use the BC at = 1 to nd C2. 0 = , 1 + C2 therefore C2 = 1 4 4 The dimensionless steady-state solution is 2 = 1 , 4; 0 1 4 The solution of th...

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