Page 62 - Fundamentals of Computational Geoscience Numerical Methods and Algorithms
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48              3  Algorithm for Simulating Coupled Problems in Hydrothermal Systems

                                             c 2 − b 2 ξ A
                                        η A =        .                   (3.58)
                                             b 3 + b 4 ξ A
            (3) a 2  = 0, a 3  = 0

                                                 2
                                       −bb ±   bb − (4aa)cc
                                  η A =                    ,             (3.59)
                                               2aa
                                             c 1 − a 3 η A
                                        ξ A =        ,                   (3.60)
                                                a 2
            where

                                             b 4 a 3
                                        aa =     ,                       (3.61)
                                              a 2
                                        b 2 a 3 − b 4 c 1
                                   bb =           − b 3 ,                (3.62)
                                            a 2
                                            −b 2 c 1
                                        cc =      .                      (3.63)
                                              a 2

            3.2.2.3 Case 3: a 4  = 0, b 4 = 0
            This case is very similar to case 2. The corresponding equations in this case are as
            follows:

                                 a 2 ξ A + a 3 η A + a 4 ξ A η A = c 1 ,  (3.64)

                                     b 2 ξ A + b 3 η A = c 2 .           (3.65)
              Similarly, the solution to Equations (3.64) and (3.65) can be expressed in the
            following three sub-cases:

            (1) b 2 = 0, b 3  = 0

                                                c 2
                                           η A =  ,                      (3.66)
                                                b 3
                                             c 1 − a 3 η A
                                        ξ A =        .                   (3.67)
                                             a 2 + a 4 η A
            (2) b 2  = 0, b 3 = 0

                                                c 2
                                           ξ A =  ,                      (3.68)
                                                b 2
                                             c 1 − a 2 ξ A
                                        η A =        .                   (3.69)
                                             a 3 + a 4 ξ A
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