Page 49 - Design and Operation of Heat Exchangers and their Networks
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36    Design and operation of heat exchangers and their networks


                           t² c





                    t¢ h                                        t² h
                    .
                    C h



                            t¢ h                      .
                                        Dt            C c  t¢ c
                            t c ²      q
                                                         t² h
                                                         t¢ c
                                                             z
                    (A)      0                          L
                               .
                            t¢ c  C c





                    t¢ h                                        t² h
                    .
                    C h



                            ¢
                           t h
                                       Dt                t² c
                                                         t²
                                      q                  h
                                                         t² c
                            ¢
                            t c
                                                             z
                    (B)      0                          L
          Fig. 2.7 Counterflow (A) and parallel-flow (B) arrangements.

                                       Z  A
                                     1
                                           ð t h  t c ÞdA
                                     A
                  Δt m                  0
            F ¼        ¼                                           1 (2.76)
                                                  0
                                                     00
                                                          00
                           0
                                     00
                Δt LM, cf  t  t 00    t  t 0  = ln t  t = t  t  0
                           h   c     h   c       h   c    h   c
             Then, Eq. (2.68) can be expressed as

                                                        00
                                              0
                                              t  t 00 c    t  t 0 c
                                                        h
                                              h
                      Q ¼ FkAΔt LM, cf ¼ FkA                          (2.77)
                                                0
                                            ln t  t = t  t  0
                                                         00
                                                    00
                                                h   c   h   c
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