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CSUN ME 501B - Homework Solutions

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College of Engineering and Computer ScienceMechanical Engineering DepartmentMechanical Engineering 501BSeminar in Engineering AnalysisSpring 2009 Class: 14443 Instructor: Larry CarettoMarch 30 Homework Solutions1. Hoffman, page 582, problem 7. Solve the heat diffusion equation problem presented in Section 9.1 using the five-point method with x = y = 5 cm using Gauss elimination. Compare the results with the exact solution in Table 9.1.The horizontal and vertical dimensions of the region are 10 cm and 15 cm, respectively. Thus the grid with a spacing of x = y = 5 cm has only two nodes in the region that do not have their values specified by boundary conditions. These are noted as T1 and T2 in the diagram to the left.The general five-point equation related the central node to its four nearest neighbors. For x = y, the central node is the average of its four nearest neighbors.41111 ijjijiijijuuuuuApplying this formula to the two unknown nodes in this problem gives the following two equations.41000040001221TTandTTThese two equations give 4T1 = T2 and 4T2 = T1 + 100. Combining the two equations to eliminate T2 gives 16T1 = T1 + 100 or T1 = 100/15= 6.67 and T2 = 4 T1 = 400/15 = 26.67. This compares to exact values of 4.13 and 20.75 shown in Table 9.1.2. Repeat the solution to problem 9.1 with x = y = 2.5 cm using SOR where the relaxation factor is found form equation 9.51.The grid for this problem is shown on the next page. Here there are 15 unknown temperature values in the center of the grid. Each of these is found by the SOR iteration relationship for the five point star:  014)1()(1)(1)1(1)1(1)1( nijnijnjinjinijnijuuuuuuThe grid for this problem is shown on the next page. Because of symmetry, the grid locations at i = 3 have the same values as those at i = 1. Thus we can save time by setting all values of T3j equal to the corresponding values of T1j. In this way, we would not have to solve for values of T3j and would have only ten unknown values of T. Jacaranda (Engineering) Room 3333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.706200100000T2T1The zero-boundary conditions at the left, bottom and top sides give the following temperature values: T10 = T20 = T30 = T01 = T41 = T02 = T42 = T03 = T43 = T05 = T45 = 0. At the top row, the boundary condition that T(x,H) = 100 sin (x/L) gives the following temperature values: T16 = T36 = 100 sin (/4) = 79.71068 andT26 = 100 sin(/2) = 100.We can use these boundary values to write the finite-difference equations for the unknown values of Tij at the nodes in the grid as shown below. We start with the two j = 1 nodes.  0125.0)(11)(12)(21)1(01)1(10)1(11 nnnnnnTTTTTT  0125.0)(21)(22)(31)1(11)1(20)1(21 nnnnnnTTTTTTThe value of T31 is the same as T11, by symmetry. Applying the boundary and symmetry conditions gives these first two equations as follows.  0125.0)(11)(12)(21)1(11 nnnnTTTT  01225.0)(21)(22)1(11)1(21 nnnnTTTTWe have a similar pair of equations for each row from j = 2 to j = 5 with slight differences due to the differences in the boundary conditions. Rows j = 2 to j = 4 have only an east and west boundary condition; row j = 5 has a north boundary condition. The equations for these rows are as follows.  0125.0)(12)(13)(22)1(11)1(12 nnnnnTTTTT  01225.0)(22)(23)1(12)1(21)1(22 nnnnnTTTTT  0125.0)(13)(14)(23)1(12)1(13 nnnnnTTTTT  01225.0)(23)(24)1(13)1(22)1(23 nnnnnTTTTT  0125.0)(14)(15)(24)1(13)1(14 nnnnnTTTTT  01225.0)(24)(25)1(14)1(23)1(24 nnnnnTTTTT  )(12)(25)1(14)1(15171.7925.0nnnnTTTT  )(22)1(15)1(24)1(251100225.0nnnnTTTTJacaranda (Engineering) Room 3333 Mail Code Phone: 818.677.6448E-mail: [email protected] 8348 Fax: 818.677.7062i = 0 1 2 3 4j =6543210Equation 9.51 gives the relaxation factor as 112opt, where  is given by equation 9.52 as    2221coscosJI. In this equation  is the step size ratio, which is one in this problem, and I and J are the number of grid spacings in the x and y direction which are 4 and 6, respectively. This gives a value of  = 0.618686 and opt = 1.236741. To start the iterations weset all the temperatures to zero and compute the temperatures from the equations above in the order shown. The first iteration gives all zeros until we reach the final row. Here we get the following results using  = 1.236741.    86.21)0(171.790025.0110025.0)(12)(25)1(14)1(15nnnnTTTT      425.44)0(110086.21225.01100225.0)(22)1(15)1(24)1(25nnnnTTTTNote the use of T15 just computed in the calculation of T25. The results of 20 iterations are shown in the table below. The exact solutions, copied from the text, are shown in the final row. At the final iteration step, all values agree to at least four significant figures. The agreement with the exact solution is not good, however, because of the small grid size.SOR Results for each iteration (n) at grid cells (i,j)n 1,1 2,1 1,2 2,2 1,3 2,3 1,4 2,4 1,5 2,50 0 0 0 0 0 0 0 0 0 01 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 21.85793 44.425142 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 6.75668 17.90982 32.51037 46.041833 0.00000 0.00000 0.00000 0.00000 2.08861 6.82750 14.63365 21.15473 32.92603 46.919564 0.00000 0.00000 0.64563 2.50965 6.33970 9.62001 15.21661 21.88235 33.27927 47.155315 0.19957 0.77578 2.64451 4.25499 6.99575 10.12968 15.61566 22.18742 33.39196 47.263546 1.01008 1.25523 3.16469 4.46962 7.28231 10.34697 15.73902 22.29217 33.43690 47.298107 1.12742 1.70928 3.23288 4.66855 7.34093 10.42570 15.77423 22.32419 33.44785 47.306608 1.26111 1.73595 3.33769 4.71888 7.39469 10.46578


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