Page 165 - Handbook of Energy Engineering Calculations
P. 165

occurs when the vortex shedding frequency is close to the acoustic frequency.
               Excessive noise is generated, leading to large gas pressure drops and bundle
               and  casing  damage.  The  starting  point  in  the  evaluation  for  noise  and
               vibration is the estimation of various frequencies.
                  Use  the  listing  of  C  values  shown  below  to  determine  the  mode  of

               vibration. Note that C is a factor determined by the end conditions of the tube
               bundle.













               Since  the  tubes  are  fixed  at  both  ends,  i.e.,  clamped,  select  the  mode  of
               vibration  as  1,  with  C  =  22.37.  For  most  situations,  Mode  1  is  the  most
               important case.



               2. Find the natural frequency of the tube bundle
                                                        4
                                                 4
                                                                     0.5
                                                              2
               Use the relation, f  = 90 C[d  – d ]/(L  – M ). Substituting, with C = 22.37,
                                                  o
                                     n
                                                         i
                                                             2
                                                 4 0.5
                                      4
                                                                       0.5
               f   =  (90)(22.37)[2   –  1.84 ] /(13.5   −  1.67 )  =  18.2  cycles  per  second
                n
               (cps). In Mode 2, f  = 50.2, as C = 61.67.
                                      n
               3. Compute the vortex shedding frequency
               To compute the vortex shedding frequency we must know several factors, the
               first  of  which  is  the  Strouhl  number,  S.  Using  Fig.  19  with  a  transverse
               pitch/diameter of 1.75 and a longitudinal pitch diameter of 1.5, we find S =
                                                                                     3
                                                                                                    3
               0.33. Then, the air density = 40/(460 − 219) = 0.059 lb/ft  (0.95 kg/m ); free
                                                                      2
                                                                                  2
               gas  area  =  40(3.5  −  2)(13.5/12)  =  67.5  ft   (6.3  m );  gas  velocity,  V  =
               300,000/(67.5)(0.059)(3600) = 21 ft/s (6.4 m/s).
                  Use the relation, f  = 12(S)(V)/d  = 12(0.33)(21)/2 = 41.6 cps, where f  =
                                                                                                         c
                                        c
                                                          o
               vortex shedding frequency, cps.
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