1 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. CHAPTER 27 27.1 The solution can be assumed to be T = ex. This, along with the second derivative T” = 2ex, can be substituted into the differential equation to give 20.15 0xxee which can be used to solve for 20.15 00.15 Therefore, the general solution is 0.15 0.15 xxTAe Be The constants can be evaluated by substituting each of the boundary conditions to generate two equations with two unknowns, 240150 48.08563 0.020796ABAB which can be solved for A = 3.016944 and B = 236.9831. The final solution is, therefore, 0.15 0.15 3.016944 236.9831xxTe e which can be used to generate the values below: x T 0 240 1 165.329 2 115.7689 3 83.79237 4 64.54254 5 55.09572 6 54.01709 7 61.1428 8 77.55515 9 105.7469 10 150 08016024002468102 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. 27.2 Reexpress the second-order equation as a pair of ODEs: 0.15dTzdxdzTdx 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 T z k11 k12 Tend zend k21 k22 1 2 0 240.000 -120.000 -120.000 36.000 228.000 -116.400 -116.400 34.200 -118.200 35.100 0.1 228.180 -116.490 -116.490 34.227 216.531 -113.067 -113.067 32.480 -114.779 33.353 0.2 216.702 -113.155 -113.155 32.505 205.387 -109.904 -109.904 30.808 -111.529 31.657 0.3 205.549 -109.989 -109.989 30.832 194.550 -106.906 -106.906 29.183 -108.447 30.007 0.4 194.704 -106.988 -106.988 29.206 184.006 -104.068 -104.068 27.601 -105.528 28.403 0.5 184.152 -104.148 -104.148 27.623 173.737 -101.386 -101.386 26.061 -102.767 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, 60 120(0) 120 (150 ( 1671.817)) 90.61262047.766 ( 1671.817)z The resulting fit, along with the two “shots” are displayed below: -2000-100001000200030000246810 The final shot along with the analytical solution (displayed as filled circles) shows close agreement: 05010015020002468103 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. 27.3 A centered finite difference can be substituted for the second derivative to give, 11220.15 0iiiiTTTTh or for h = 1, 112.15 0iiiTTT The first node would be 122.15 240TT and the last node would be 9102.15 150TT The tridiagonal system can be solved with the Thomas algorithm or Gauss-Seidel for (the analytical solution is also included) x T Analytical 0 240 240 1 165.7573 165.3290 2 116.3782 115.7689 3 84.4558 83.7924 4 65.2018 64.5425 5 55.7281 55.0957 6 54.6136 54.0171 7 61.6911 61.1428 8 78.0223 77.5552 9 106.0569 105.7469 10 150 150 The following plot of the results (with the analytical shown as filled circles) indicates close agreement. 0801602400246810 27.4 The second-order ODE can be expressed as the following pair of first-order ODEs, 27dyzdxdz z y xdx4 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. 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 z(0) 1 0.5 y(20) 11,837.64486 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, 0.5 1(0) 1 (8 ( 11837.64486)) 0.8285723922712.34615 ( 11837.64486)z The resulting fit is displayed below: x y 0 5 2 4.151601 4 4.461229 6 5.456047 8 6.852243 10 8.471474 12 10.17813 14 11.80277 16 12.97942 18 12.69896 20 8 048120 5 10 15 20 27.5 Centered finite differences can be substituted for the second and first derivatives to give, 11112272 02iii iiiiyyy yyyxxx or substituting x = 2 and collecting terms yields 112.25 4.5 1.25iiiiyyyx 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 y 0 55 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
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