Page 63 - Introduction to Computational Fluid Dynamics
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                            Table 2.5: Coefficients in the discretised          1D HEAT CONDUCTION
                            equation – Problem 2.
                            i     2         3      4      5      6
                            AW i     0      4.5    4.5    4.5    4.5
                            AE i     4.5    4.5    4.5    4.5    0
                                  2025.6    0.6    0.6    0.6    0.6
                            Su i
                                     9.024  0.024  0.024  0.024  0.024
                            Sp i
                            used. Greater accuracy can be obtained with finer grids; however, this will require
                            more computational effort.
                               From the converged solution, the fin heat loss is estimated as Q loss = AW 2 ×
                            (T 1 − T 2 ) = 9(225 − 222.42) = 23.26 W. This also compares favourably with the
                            exact solution already mentioned.
                               Table 2.7 shows the execution of the same problem using TDMA. The table
                            shows values of A i and B i derived fromTable2.5and Equations2.74 and 2.75. Since
                            these are constants, solution is now obtained in only one iteration. Also, the initial
                            guess becomes irrelevant. The estimated heat loss is Q loss = 9(225 − 222.45) =
                            22.967 W.
                               Thus, compared to GS, the TDMA procedure is considerably faster. Experience
                            shows that this conclusion is valid even in nonlinear problems. For this reason, the
                            TDMA is the most preferred solution procedure in generalised codes.


                            Problem 3 – Annular Composite Fin
                            Consider an annular fin put on a tube (of outer radius r 1 = 1.25 cm), as shown
                            in Figure 2.11. The fin is made from two materials: The inner material has radius
                            r 2 = 2.5 cm and conductivity k 2 = 200 W/m-K and the outer material extends
                            to radius r 3 = 3.75 cm and has conductivity k 3 = 40 W/m-K. The fin thickness
                                                                                             ◦
                            t = 1 mm. The tube wall (and hence the fin base) temperature is T 0 = 200 C. The

                            Table 2.6: Solution by Gauss–Seidel method – Problem 2.

                            l      FCMX      0 cm   0.2 cm  0.6 cm  1.0 cm  1.4 cm  1.8 cm  2.0 cm
                             0               225    223     219     215     211     207     205
                             1     0.01      225    222.65  218.31  214.15  210.08  209.1   209.1
                             2     0.0034    225    222.42  217.77  213.44  210.77  209.78  209.78
                             3     0.0021    225    222.24  217.32  213.54  211.16  210.18  210.18
                             .      .         .        .       .       .       .       .       .
                             .      .         .        .       .       .       .       .       .
                             .      .         .        .       .       .       .       .       .
                            22     0.00012   225    222.41  218.28  215.22  213.19  212.19  212.19
                            23     0.00011   225    222.41  218.30  215.24  213.21  212.21  212.21
                            24     0.000092  225    222.42  218.31  215.25  213.23  212.23  212.23
                            Exact     −      225    222.58  218.52  215.51  213.49  212.49  212.37
   58   59   60   61   62   63   64   65   66   67   68