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114    Chapter 4 Shell and tube heat exchanger




                In order to calculate bundle cross flow area A e , we adopt B ¼ 0:5D s ¼ 0:5   0:489 ¼ 0:2445 m,
             which is less than the maximum unsupported length as per Table 4.13 and is greater than 50 mm or
             D s =5 (as discussed in Section 4.6.6). and obtain A e from Eq. (4.10) as
                         D s  ðP T   D 0 Þ  B  0:489  ð0.03175   0:0254Þ  0:2445      2
                                                                            ¼ 0:0239 m
                     A e ¼                 ¼
                                 P T                     0.03175
                From Eq. (4.9), shell-side mass velocity is
                                           m s  10:8472             2
                                                      ¼ 453:858 kg=m s
                                      G s ¼  ¼
                                           A e  0:0239
             and the linear velocity on the shell side u s is
                     G s  453:858
                                 ¼ 0:6392 m s, which is > 0.6 m/s and <1.5 m/s (Ok as per point 8 in
                u s ¼   ¼
                     r s    710
             Section 4.4).
                In order to estimate h o , shell-side equivalent diameter De is calculated for triangular pitch from
             Eq. 4.12(b) as

                                                n                               o
                                                                             2
                                                        2
                           2          2      4   0:03175   0:866   p  ð0:0254Þ =4
                    4   P sin 60   pD =4

                          T           o
                De ¼                      ¼                                       ¼ 0:01836 m
                             pD o                         p   0:0254
                                        0:01836   453:858
                                DeG s
                                                        ¼ 30862:3
                                 m s        0:00027
                This gives Re De ¼   ¼
                                           6
                           3
                Since 2   10 < Re De < 1   10 , we obtain Nu De from Eq. (4.11) as
                                         1
                                     0:55                  0:55        0:333
                         Nu De ¼ 0:36Re  3                                 ¼ 172:052
                                     De  Pr s ¼ 0:36  ð30862:3Þ   ð4:2715Þ
                                           Nu De   k s  172:052   0:148225        W
                                                                       ¼ 1389:02     .
                And h o from Eq. (4.13) as h o ¼    ¼                             2
                                              De           0:01836               m K
                To obtain h i , we use correlation from Table 2.6 i.e., Nu ¼ 0:023   Re 0:8   ðPrÞ 0:33  where Re t ¼
             D i u t r t  0:01859   1:985   992:22  ¼ 57031:24.
               m t  ¼       0:000642
                This gives Nu t ¼ 0:023   Re 0:8    Pr 0:333          0:8       0:33  ¼ 236:768.
                                        t      t   ¼ 0:023  ð57031:24Þ    ð4:26Þ
                                    236:768   0:62976
                          Nu i   k i                          W
                                                    ¼ 8020:82  2 .
                and h i ¼¼        ¼                          m K
                            D i         0:01859
                In order to calculate tube wall resistance R w , we adopt carbon steel (1.5%C) tubes from Table 4.9
             and obtain its thermal conductivity k w from literature as 36 W/mK.
                Considering unfinned tubes, R w from Eq. (4.15a) is
                                                                               2
                                 D o     D o      0:0254   0:0254          4  m K
                                    ln                  ln       ¼ 1:1   10
                           R w ¼                ¼
                                2k w   D o   2t   2   36  0:01859             W
                Based on fouling factor Table E1A included in Appendix E, for treated cooling water with inhibitor
                                                          2                              2
             for velocity>1 m/s, typical fouling factor is 0.00018 m K/W and for kerosene it is 0.00018 m K/W.
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