Tema27 - CHAPTER 27 27.1 The solution can be assumed to be...

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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 second-order 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 step-size 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 0 200.000-55.000-55.000 20.000 194.500-53.000-53.000 19.450-54.000 19.725 0.1 194.600-53.028-53.028 19.460 189.297-51.082-51.082 18.930-52.055 19.195 0.2 189.395-51.108-51.108 18.939 184.284-49.214-49.214 18.428-50.161 18.684 0.3 184.378-49.240-49.240 18.438 179.454-47.396-47.396 17.945-48.318 18.192 0.4 179.547-47.420-47.420 17.955 174.805-45.625-45.625 17.480-46.523 17.718 0.5 174.894-45.649-45.649 17.489 170.330-43.900-43.900 17.033-44.774 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:-300-200-100 100 200 300 400 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 Gauss-Seidel 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 second-order ODE can be expressed as the following pair of first-order 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.

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Tema27 - CHAPTER 27 27.1 The solution can be assumed to be...

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