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Unformatted text preview: CHAPTER 27 27.1 The solution can be assumed to be T = e λ x . This, along with the second derivative T ” = λ 2 e λ x , can be substituted into the differential equation to give λ λ λ 2 01 e e x x = . which can be used to solve for λ λ 2 01 01 = = ± . . Therefore, the general solution is T Ae Be x x = + 0.1 0.1 The constants can be evaluated by substituting each of the boundary conditions to generate two equations with two unknowns, 200 100 23 62434 0 042329 = + = + A B A B . . which can be solved for A = 3.881524 and B = 196.1185. The final solution is, therefore, T e e x x = + 3881524 1961185 0.1 0.1 . . which can be used to generate the values below: x T 200 1 148.2747 2 111.5008 3 85.97028 4 69.10864 5 59.21565 6 55.29373 7 56.94741 8 64.34346 9 78.22764 10 100 100 200 2 4 6 8 10 27.2 Reexpress the secondorder equation as a pair of ODEs: dT dx z dz dx T = = 01 . The solution was then generated on the Excel spreadsheet using the Heun method (without iteration) with a stepsize of 0.01. An initial condition of z =  55 was chosen for the first shot. The first few calculation results are shown below. x T z k11 k12 Tend zend k21 k22 phi1 phi2 0 200.00055.00055.000 20.000 194.50053.00053.000 19.45054.000 19.725 0.1 194.60053.02853.028 19.460 189.29751.08251.082 18.93052.055 19.195 0.2 189.39551.10851.108 18.939 184.28449.21449.214 18.42850.161 18.684 0.3 184.37849.24049.240 18.438 179.45447.39647.396 17.94548.318 18.192 0.4 179.54747.42047.420 17.955 174.80545.62545.625 17.48046.523 17.718 0.5 174.89445.64945.649 17.489 170.33043.90043.900 17.03344.774 17.261 The resulting value at x = 10 was T (10) = 315.759. A second shot using an initial condition of z (0) =  70 was attempted with the result at x = 10 of T (10) =  243.249. These values can then be used to derive the correct initial condition, z ( ) . . ( . ) . 55 70 55 243 249 315 759 100 315 759 60 79 =  + + =  The resulting fit, along with the two “shots” are displayed below:300200100 100 200 300 400 2 4 6 8 10 27.3 A centered finite difference can be substituted for the second derivative to give, T T T h T i i i i + + = 1 1 2 2 01 . or for h = 1, + = + T T T i i i 1 1 21 . The first node would be 21 200 1 2 . T T = and the last node would be + = T T 9 10 21 100 . The tridiagonal system can be solved with the Thomas algorithm or GaussSeidel for (the analytical solution is also included) x T Analytical 200 200 1 148.4838 148.2747 2 111.816 111.5008 3 86.32978 85.97028 4 69.47655 69.10864 5 59.57097 59.21565 6 55.62249 55.29373 7 57.23625 56.94741 8 64.57365 64.34346 9 78.3684 78.22764 10 100 100 27.4 The secondorder ODE can be expressed as the following pair of firstorder ODEs, dy dx z dz dx z y x = = + 2 8 These can be solved for two guesses for the initial condition of z . For our cases we used z (0) 1 0.5 y (20) 6523.000507 7935.937904 Clearly, the solution is quite sensitive to the initial conditions. These values can then be used toClearly, the solution is quite sensitive to the initial conditions....
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This note was uploaded on 03/13/2012 for the course AEROSPACE 301 taught by Professor Pfchang during the Spring '12 term at Shandong University.
 Spring '12
 pfchang

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