Page 111 - Design and Operation of Heat Exchangers and their Networks
P. 111

Steady-state characteristics of heat exchangers  99

                                 ∞
                                X    k        k +1Þ=2     p ffiffiffiffiffi
                  G n x, yÞ ¼ e  x y    ð y=xÞ ð   I k +1 2 xy n   0ð  Þ  (3.173)
                    ð
                                     n
                                k¼n
                                  ∞            k            j
                                 X    y k + n +1 X  ð k + n jÞ!x
                              x y
                  G n x, yð  Þ ¼ e                           ð n   0Þ   (3.174)
                                    ð k + n +1Þ!   ð k jÞ!n!j!
                                 k¼0           j¼0
              to calculate F n (x, y) and G n (x, y).
              3.3.4.3 Examples for cross counterflow arrangements
              The ε-NTU relationships for the cross counterflow arrangement types with
              at least one fluid unmixed throughout can be found in Table 3 of Baclic
              (1990). We represent the ε-NTU relationship examples for the cross
              counterflow arrangements BA m,5 (m¼1, 2, …, 5) and BA m,6 (m¼1, 2, …,6)
              as follows. For the flow arrangement BA m,n (m>n), we can exchange
              the fluid indices “1” and “2” and find the corresponding flow arrangement
              listed in the succeeding text. The same symbols of Baclic are used in the
              expressions

                                 a A ¼ NTU 1,A , a B ¼ NTU 1,B          (3.175)
                                                                        (3.176)
                  b A ¼ NTU 2,A ¼ R 1 NTU 1,A , b B ¼ NTU 2,B ¼ R 1 NTU 1,B
                           a ¼ NTU 1 =2, b ¼ NTU 2 =2 ¼ R 1 NTU 1 =2    (3.177)

                                        NTU 1,B  a B  b B
                                    ϕ ¼        ¼    ¼                   (3.178)
                                        NTU 1,A  a A  b A
                                                                        (3.179)
                                 NTU 1 ¼ NTU 1,A + NTU 1,B
                 Let R 1 ¼2, NTU 1,A ¼0.4, NTU 1,B ¼0.6; we have

                  a A ¼ 0:4, b A ¼ 0:8, a B ¼ 0:6, b B ¼ 1:2, ϕ ¼ 1:5, NTU 1 ¼ 1
                                 1             1       0:8
                        Kb A ¼     1 e   b A  ¼   1 e      ¼ 0:6883,
                          ðÞ
                                b A           0:8
                                 1             1       1:2
                          ðÞ
                        Kb B ¼     1 e   b B  ¼   1 e      ¼ 0:5823,
                                b B           1:2
                                                          ∗
                  ∗
                                                                        ðÞ
                                ðÞ
                 ν a A b A Þ ¼ e  a A Kb A  ¼ e  0:4 0:6883  ¼ 0:7593, ν a B b B Þ ¼ e  a B Kb B
                                                           ð
                   ð
                             0:6 0:5823
                         ¼ e         ¼ 0:7051:
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