Page 118 - Design and Operation of Heat Exchangers and their Networks
P. 118

106   Design and operation of heat exchangers and their networks



             Example BA 6,6 —cont’d

                For the data given in Example BA 5,6 , we have
                               "                         #
                                 id
                                                ðÞ
                                A ð b A , b B , a A Þ B mn b B
                                 mn
                           A ¼
                                              id
                                   ðÞ
                                B mn a A     A ð a B , a A , b B Þ
                                              mn
                               2                             3
                                0:24212 0:05505   1      0:6
                                0:21592 0:04469   1      0:8
                               6                             7
                             ¼  6                            7
                               4   1      0:2  0:59887 0:41189  5
                                   1    0:26667 0:56695 0:38401
             and get the inverse matrix as
                          2                                     3
                            0:36246  0:44106   4:16148   2:97849
                    A  1  ¼  6   0:1378  0:68478  15:1034  14:9889  7
                          6
                                                                7
                          4  4:12564  2:90520  0:30562  0:065989  5
                            5:05152   4:96229  0:10252  0:049083
             which yields
                              2  3      2 3   2       3
                               α 0       1      0:37947
                              6  7     1 1    6 0:43249  7
                                        6 7
                               α 1
                              6  7  ¼ A  6 7  ¼  6    7
                               β
                                         1
                              4  5      4 5   4 0:84883  5
                                0
                               β  1      1      0:06238
                      "                    #
                             N
                    1      1  X
                ε 1 ¼  1       α n F n +1 b A , a A Þ
                                     ð
                    R 1   a A
                            n¼0

                    1     1
                  ¼   1    ð 0:37947 0:20630 + 0:43249 0:03956Þ ¼ 0:3808
                    2    0:4
          3.3.4.4 Examples for cross parallelflow arrangements
          The ε-NTU relationships for the cross parallelflow arrangement types with
          at least one fluid unmixed throughout can be found in Table 5 of Baclic
          (1990), in which the two passes are equally sized, that is, ϕ¼1. Totally,
          11 examples of such flow arrangements are represented as follows:
             Let R 1 ¼2 and NTU 1 ¼1; we have
                                                            1       b
               a ¼ NTU 1 =2 ¼ 0:5, b ¼ R 1 NTU 1 =2 ¼ 1, KbðÞ ¼  1 e
                                                            b
                   1      1
                 ¼   1 e    ¼ 0:6321,
                   1
   113   114   115   116   117   118   119   120   121   122   123