Page 94 - Design and Operation of Heat Exchangers and their Networks
P. 94

82    Design and operation of heat exchangers and their networks


          Then, we have
                                     1 e  NTU h 1+ R h Þ
                                               ð
                                 ε h ¼                                (3.62)
                                          1+ R h
                                     1 e  NTU c 1+ R c Þ
                                               ð
                                 ε c ¼                                (3.63)
                                          1+ R c
          where
                                                _
                                   NTU h ¼ kA=C h                     (3.64)
                                           _
                              NTU c ¼ kA=C c ¼ R h NTU h              (3.65)
                                           _  _
                                     R h ¼ C h =C c                   (3.66)
                                       _   _
                                  R c ¼ C c =C h ¼ 1=R h              (3.67)
             The relation between the two dimensionless temperature changes is


                                      ε c ¼ R h ε h                   (3.68)
             Furthermore, since “hot” and “cold” is only a relative description, there-
          fore, we can refer the “hot” fluid to any one of the two fluids and the “cold”
          fluid to the other.



          3.2.2.2 Counterflow heat exchangers
          For a counterflow heat exchanger, the ε-NTU relation is given by
          Eq. (3.37) as

                                         ð
                                  ð
                 t  t c 0    1 e   kA= _ C cÞ 1  _ C c = _ C hÞ  1 e  NTU c 1 R c Þ
                  00
                                                                  ð
                  c
             ε c ¼     ¼                             ¼
                 t  t 0        _  _       kA= _ C cÞ 1  _ C c = _ C hÞ  1 R c e  NTU c 1 R c Þ
                  0
                                                                   ð
                                       ð
                                             ð
                  h   c  1  C c =C h e
                                                                      (3.69)
             Using Eq. (3.68), we can obtain
                                                   ð
                                         1 e  NTU h 1 R h Þ
                            ε h ¼ ε c =R h ¼                          (3.70)
                                        1 R h e  NTU h 1 R h Þ
                                                    ð
             Because Eqs. (3.69), (3.70) have the same form, they can be expressed by
          the ε-NTU relation as
                                              ð
                                      1 e  NTU 1 RÞ
                                  ε ¼                                 (3.71)
                                     1 Re   NTU 1 RÞ
                                               ð
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