Page 86 - Design and Operation of Heat Exchangers and their Networks
P. 86

74    Design and operation of heat exchangers and their networks


                  0
          where V is named as the inlet matrix:
                                     "             #
                                      h 11 e r 1 z 0 1 h 12 e r 2 z 0 1
                                 V 5       0      0                   (3.17)
                                  0
                                      h 21 e r 1 z 2 h 22 e r 2 z 2
             The coefficient matrix D can then be expressed with the inverse matrix
          of the inlet matrix multiplied by the inlet temperature vector:

                                           0  1  0
                                     D ¼ V    T                       (3.18)
             The final solution can then be expressed as

                                          Rz
                                   T ¼ He V  0  1 T 0                 (3.19)
             Using the obtained eigenvalues and eigenvectors as well as the boundary
          condition, we can explicit express the terms in Eq. (3.19) as
                                     "                            #
                         r 1 z  r 2 z            kA= _ C h + kA= _ C cÞz
               Rz    h 11 e  h 12 e   1       e  ð
             He   ¼      r 1 z  r 2 z  ¼               kA= _ C h + kA= _ C cÞz  (3.20)
                                            _
                                                _
                     h 21 e  h 22 e   1   C h =C c e  ð
                        "      0      0  #  1  "           #  1
                          h 11 e r 1 z 1 h 12 e r 2 z 1  1  1
                   0  1
                  V   ¼        0      0    ¼         _  _
                          h 21 e r 1 z 2 h 22 e r 2 z 2  1   C h =C c
                        2   _   _                3
                            C h =C c      1
                        6     _   _       _   _  7
                        6  1+ C h =C c  1+ C h =C c 7
                                                                      (3.21)
                      ¼ 6                        7
                              1            1
                        4                        5

                              _   _        _   _
                          1+ C h =C c  1+ C h =C c
          and finally obtain the temperature distributions:
                                       _
                                   _
                                             ð
                           t h  t  0  C h =C c + e   kA= _ C h + kA= _ C cÞz
                               c
                                 ¼                                    (3.22)
                            0
                           t  t  0            _  _
                            h  c          1+ C h =C c
                                       h      kA= _ C h + kA= _ C cÞz  i
                                  _
                                      _
                                              ð
                          t c  t 0  C h =C c 1 e
                              c
                               ¼                                      (3.23)
                          t  t  0             _  _
                          0
                          h   c           1+ C h =C c
             The outlet temperatures (z ¼ 1) are
                                        _
                                    _
                                             ð
                            00
                           t  t  0  C h =C c + e   kA= _ C h + kA= _ C cÞ
                            h   c
                                 ¼                                    (3.24)
                            0
                            t  t 0            _  _
                            h   c         1+ C h =C c
   81   82   83   84   85   86   87   88   89   90   91