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 30 30.1 The key to approaching this problem is to recast the PDE as a system of ODEs. Thus, by substituting the finite-difference approximation for the spatial derivative, we arrive at the following general equation for each node 1122ii iidT T T Tkdtx By writing this equation for each node, the solution reduces to solving 4 simultaneous ODEs with Heun’s method. The results for the first two steps along with some later selected values are tabulated below. In addition, a plot similar to Fig. 30.4, is also shown t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 100 0 0 0 0 50 0.1 100 2.043923 0.021788 0.010894 1.021962 50 0.2 100 4.005178 0.084022 0.042672 2.002593 50 - - - 3 100 37.54054 10.27449 6.442319 18.95732 50 6 100 53.24294 24.66052 17.4603 27.92251 50 9 100 62.39032 36.64937 27.84901 34.34692 50 12 100 68.71331 46.03498 36.54213 39.5355 50 0501000510 30.2 Because we now have derivative boundary conditions, the boundary nodes must be simulated. For node 0, 100101(2 )ll lllTTTTT (i) This introduces an exterior node into the solution at i = 1. The derivative boundary condition can be used to eliminate this node, 1102TTdTdx x which can be solved for 0112dTTT xdx2 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. which can be substituted into Eq. (i) to give 1000 10222lll lldTTT TT xdx For our case, dT0/dx = 1 and x = 2, and therefore T1 = T1 – 4. This can be substituted into Eq. (i) to give, 100 10(224)ll llTT TT A similar analysis can be used to embed the zero derivative in the equation for the nth node, 111(2 )ll l llnn n nnTTT TT (ii) This introduces an exterior node into the solution at n + 1. The derivative boundary condition can be used to eliminate this node, 112nnnTTdTdx x which can be solved for 112nnndTTT xdx which can be substituted into Eq. (ii) to give 11222ll l lnnn n ndTTT T T xdx For our case, n = 5 and dTn/dx = 0, and therefore 155 45(2 2 )ll llTT TT Together with the equations for the interior nodes, the entire system can be solved with a step of 0.1 s. The results for some of the early steps along with some later selected values are tabulated below. In addition, a plot of the later results is also shown t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 50.0000 50.0000 50.0000 50.0000 50.0000 50.0000 0.1 49.9165 50.0000 50.0000 50.0000 50.0000 50.0000 0.2 49.8365 49.9983 50.0000 50.0000 50.0000 50.0000 0.3 49.7597 49.9949 50.0000 50.0000 50.0000 50.0000 0.4 49.6861 49.9901 49.9999 50.0000 50.0000 50.0000 0.5 49.6153 49.9840 49.9997 50.0000 50.0000 50.0000 - - - 200 5.000081 6.800074 8.200059 9.200048 9.800042 10.00004 400 -11.6988 -9.89883 -8.49883 -7.49882 -6.89881 -6.69881 600 -28.4008 -26.6008 -25.2008 -24.2008 -23.6007 -23.40073 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. 800 -45.1056 -43.3056 -41.9056 -40.9056 -40.3056 -40.1056 1000 -61.8104 -60.0104 -58.6104 -57.6104 -57.0104 -56.8104 -80-60-40-200204060024681002004006008001000 Notice what’s happening. The rod never reaches a steady state, because of the heat loss at the left end (unit gradient) and the insulated condition (zero gradient) at the right. 30.3 The solution for t = 0.1 is (as computed in Example 30.1), t x = 0 x = 2 x = 4 x = 6 x = 8 x = 100 100 0 0 0 0 50 0.1 100 2.0875 0 0 1.04375 50 0.2 100 4.087847 0.043577 0.021788 2.043923 50 For t = 0.05, it is t x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 0 100 0.000000 0.000000 0.000000 0.000000 50 0.05 100 1.043750 0.000000 0.000000 0.521875 50 0.1 100 2.065712 0.010894 0.005447 1.032856 50 0.15 100 3.066454 0.032284 0.016228 1.533227 50 0.2 100 4.046528 0.063786 0.032229 2.023265 50 To assess the differences between the results, we performed the simulation a third time using a more accurate approach (the Heun method) with a much smaller step size (t = 0.001). It was assumed that this more refined approach would yield a prediction close to true solution. These values could then be used to assess the relative errors of the two Euler solutions. The results are summarized as x = 0 x = 2 x = 4 x = 6 x = 8 x = 10 Heun (h = 0.001) 100 4.006588 0.083044 0.042377 2.003302 50 Euler (h = 0.1) 100 4.087847 0.043577 0.021788 2.043923 50 Error relative to Heun 2.0% 47.5% 48.6% 2.0% Euler (h = 0.05) 100 4.046528 0.063786 0.032229 2.023265 50 Error relative to Heun 1.0% 23.2% 23.9% 1.0% Notice, that as would be expected for Euler’s method, halving the step size approximately halves the global relative error. 30.4 The approach described in Example 30.2 must be modified to account for the zero derivative at the right hand node (i = 5). To do this, Eq. (30.8) is first written for that node as 1114565(1 2 )llllTTTT (i)4 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
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