Page 418 - Design and Operation of Heat Exchangers and their Networks
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Experimental methods for thermal performance of heat exchangers  401


              in which n and m are constants taken from the literature, for example, from
              the Dittus-Boelter correlation (Dittus and Boelter, 1930) with n¼0.8,
              m¼0.4 for heating, and m¼0.3 for cooling. We also assume that the heat
              transfer coefficients of the two fluids are uniform along their flow directions,
              so the overall heat transfer coefficient is given by
                                    1     1           1
                                      ¼      + R w +                     (8.34)
                                   kA   α 1 A 1     α 2 A 2
              The substitution of Eq. (8.33) into Eq. (8.34) yields
                                       1
                                         ¼ X=C 1 + C 2                   (8.35)
                                      kA
              with

                                                  1
                                       C 2 ¼ R w +                       (8.36)
                                                 α 2 A 2
                                                   μ    0:14
                                         d h,1      w,1
                                 X ¼                                     (8.37)
                                        n  m       μ
                                     Re Pr λ 1 A 1  1
                                        1  1
                                                                    4
                 Now, we can do a set of test runs with Re 1 varying from 10 to its max-
              imal possible value and meanwhile keep α 2 unchanged when Re 1 varies.
              Applying a linear regression to the measured data set (kA) i and X i (i¼1,
              2, …, N), the constants C 1 and C 2 can be determined by minimizing
              Eq. (8.38):
                                         "                     # 2
                                      N
                                     X        1             1
                           ð
                          fC 1 , C 2 Þ ¼            + C 2                (8.38)
                                               n
                                          C 1 Re W i       ð kAÞ
                                      i¼1      1,i             i
              where
                                          m       μ    0:14
                                        Pr λ 1 A 1  1
                                          1
                                   W ¼                                   (8.39)
                                          d h,1  μ w,1
              The minimum point can be found by setting the partial derivatives with
              respect to (1/C 1 ) and C 2 to zero:
                  "                                                   #
                                        N
                                                      N
                       N
                    1  X     1         X      1      X         1
                 2                + C 2        n                 n     ¼ 0
                                 2
                      i¼1 Re W   i      i¼1    1   i  i¼1    i   1    i
                   C 1       n              Re W         ð kAÞ Re W
                             1
                                                                         (8.40)
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