1 CHAPTER 27 27 1 The solution can be assumed to be T e x This along with the second derivative T 2e x can be substituted into the differential equation to give 2 e x 0 15e x 0 which can be used to solve for 2 0 15 0 0 15 Therefore the general solution is T Ae 0 15 x Be 0 15 x The constants can be evaluated by substituting each of the boundary conditions to generate two equations with two unknowns 240 A B 150 48 08563 A 0 020796 B which can be solved for A 3 016944 and B 236 9831 The final solution is therefore T 3 016944e 0 15 x 236 9831e 0 15 x which can be used to generate the values below x 0 1 2 3 4 5 6 7 8 9 10 T 240 165 329 115 7689 83 79237 64 54254 55 09572 54 01709 61 1428 77 55515 105 7469 150 240 160 80 0 0 2 4 6 8 10 PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 2 27 2 Reexpress the second order equation as a pair of ODEs dT z dx dz 0 15T dx 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 120 was chosen for the first shot The first few calculation results are shown below x 0 0 1 0 2 0 3 0 4 0 5 T 240 000 228 180 216 702 205 549 194 704 184 152 z 120 000 116 490 113 155 109 989 106 988 104 148 k11 k12 120 000 116 490 113 155 109 989 106 988 104 148 36 000 34 227 32 505 30 832 29 206 27 623 Tend 228 000 216 531 205 387 194 550 184 006 173 737 zend 116 400 113 067 109 904 106 906 104 068 101 386 k21 k22 1 2 116 400 113 067 109 904 106 906 104 068 101 386 34 200 32 480 30 808 29 183 27 601 26 061 118 200 114 779 111 529 108 447 105 528 102 767 35 100 33 353 31 657 30 007 28 403 26 842 The resulting value at x 10 was T 10 1671 817 A second shot using an initial condition of z 0 60 was attempted with the result at x 10 of T 10 2047 766 These values can then be used to derive the correct initial condition z 0 120 60 120 150 1671 817 90 6126 2047 766 1671 817 The resulting fit along with the two shots are displayed below 3000 2000 1000 0 1000 0 2 4 6 8 10 2000 The final shot along with the analytical solution displayed as filled circles shows close agreement 200 150 100 50 0 0 2 4 6 8 10 PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 3 27 3 A centered finite difference can be substituted for the second derivative to give Ti 1 2Ti Ti 1 h2 0 15Ti 0 or for h 1 Ti 1 2 15Ti Ti 1 0 The first node would be 2 15T1 T2 240 and the last node would be T9 2 15T10 150 The tridiagonal system can be solved with the Thomas algorithm or Gauss Seidel for the analytical solution is also included x 0 1 2 3 4 5 6 7 8 9 10 Analytical 240 165 3290 115 7689 83 7924 64 5425 55 0957 54 0171 61 1428 77 5552 105 7469 150 T 240 165 7573 116 3782 84 4558 65 2018 55 7281 54 6136 61 6911 78 0223 106 0569 150 The following plot of the results with the analytical shown as filled circles indicates close agreement 240 160 80 0 0 2 4 6 8 10 27 4 The second order ODE can be expressed as the following pair of first order ODEs dy z dx dz 2 z y x 7 dx PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 4 These can be solved for two guesses for the initial condition of z For our cases we used 1 and 0 5 We solved the ODEs with the Heun method without iteration using a step size of 0 125 The results are 1 11 837 64486 z 0 y 20 0 5 22 712 34615 Clearly the solution is quite sensitive to the initial conditions These values can then be used to derive the correct initial condition z 0 1 0 5 1 8 11837 64486 0 82857239 22712 34615 11837 64486 The resulting fit is displayed below x y 0 2 4 6 8 10 12 14 16 18 20 5 4 151601 4 461229 5 456047 6 852243 8 471474 10 17813 11 80277 12 97942 12 69896 8 12 8 4 0 0 5 10 15 20 27 5 Centered finite differences can be substituted for the second and first derivatives to give 7 yi 1 2 yi yi 1 x 2 2 yi 1 yi 1 yi xi 0 2 x or substituting x 2 and collecting terms yields 2 25 yi 1 4 5 yi 1 25 yi 1 xi This equation can be written for each node and solved with methods such as the Tridiagonal solver the Gauss Seidel method or LU Decomposition The following solution was computed using Excel s Minverse and Mmult functions x 0 y 5 PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 5 2 4 6 8 10 12 14 16 18 20 4 199592 4 518531 5 …
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