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TAMU PETE 301 - Numerical Methods for Engineers Ch. 31 Solutions

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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 31 31.1 The equation to be solved is 2215dTdx Assume a solution of the form T = ax2 + bx + c which can be differentiated twice to give T” = 2a. Substituting this result into the differential equation gives a = 7.5. The boundary conditions can be used to evaluate the remaining coefficients. For the first condition at x = 0, 275 7.5(0) (0)bc   or c = 75. Similarly, for the second condition 2150 7.5(10) (10) 75b   which can be solved for b = 82.5. Therefore, the final solution is 27.5 82.5 75Txx   The results are plotted as 01002003000246810 31.2 The heat source term in the first row of Eq. (31.26) can be evaluated by substituting Eq. (31.3) and integrating to give 2.5 02.515 18.752.5xdx Similarly, Eq. (31.4) can be substituted into the heat source term of the second row of Eq. (31.26), which can also be integrated to yield 2.5 0015 18.752.5xdx These results along with the other parameter values can be substituted into Eq. (31.26) to give 12 10.4 0.4 ( ) 18.75dTTT xdx  and2 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. 12 20.4 0.4 ( ) 18.75dTTT xdx   31.3 In a manner similar to Fig. 31.7, the equations can be assembled for the total system, 1123455( ) / 18.750.4 0.437.50.4 0.8 0.437.50.4 0.8 0.437.50.4 0.8 0.4( ) / 18.750.4 0.4TdT x dxTTTdT x dxT     The known end temperatures can be substituted to give 12345()/1 0.4 11.250.8 0.4 67.50.4 0.8 0.4 37.50.4 0.8 97.50.4 1 41.25()/dT x dxTTTdT x dx These equations can be solved for 12345()/82.5234.375300271.87567.5()/dT x dxTTTdT x dx The solution, along with the analytical solution (dashed line) is shown below: 01002003000246810 31.4 220dc dcDUkcdxdx 22dc dcRD U kcdxdx 212 2 xixdc dcDUkcNdxdxdx3 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. 212 2 () xixdcDNxdxdx (1) 21 () xixdcUNxdxdx (2) 21 () xixkcNxdx (3) Term (1): 211221 212 21221()() ()xixccdcxdcdx x xDNxdxDccdcdxxdx x x Term (2): 2211 21 21() () xxiixxccdcN x dx N x dxdx x x 21 21 () 2xixxxNxdx 21 21 () 2xixccdcNxdxdx 2121 212() 2xixccdcUNxdxUccdx Term (3): 21 211 2()() 2xixkx xckcNxdxc Total element equation [(1) + (2) + (3)] 212122211211bbccaaaa where 11 2 12122DUkaxxxx 12212DUaxx 21212DUaxx 22 2 12122DUkaxxxx 11()dcbDxdx 22()dcbD xdx 31.5 First we can develop an analytical function for comparison. Substituting parameters gives4 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. 2920.1 200 10 1000dudx Assume a solution of the form cbxaxu 2 This can be differentiated twice to yield d2u/dx2 = 2a. This can be substituted into the ODE, which can then be solved for a = 2.5108. The boundary conditions can then be used to evaluate the remaining coefficients. At the left side, u(0) = 0 and cb )0()0(105.2028 and therefore, c = 0. At the right side of the bar, u(10) = 0 and )10()10(105.2028b and therefore, b = 2.5107, and the solution is xxu728105.2105.2 which can be displayed as -8.E-07-6.E-07-4.E-07-2.E-070.E+000246810 The element equation can be written as 2121)()()()()()(111121212112xxxxccdxxNxPdxxNxPxdxduxdxduEAuuxxEA The distributed load can be evaluated as 1000201000 1000221000 2 0 2 0 dxxdxx Thus, the final element equation is5 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.


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