Page 125 - Modelling in Transport Phenomena A Conceptual Approach
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4.5.  FLOW IN CIRCULAR PIPES                                        105

           In problems dealing  with the flow of  compressible fluids, it is customary to define
           mass velocity, G, as
                                             m
                                             A
                                         G=-=p(  v)                             (2)
           The advantage of  wing  G is the fact  that  it remains constant for steady flow  of
           compressible fluids through ducts of uniform cross-section. In this case
                                G = (1.2047)(50) = 60.24 kg/ m2. s

            The inventory rate equation for energy is written as
                              Rate of  energy in = Rate  of  energy out         (3)
           Equations  (3)-(5) of Example 4.15 are  also  applicable to this problem.  Therefore,
           we get                 -- f E ( Tw - Tbi, )
                                  L
                                              In
                                  D-4  Nu        Tw - Gout                      (4)
           The Nzlsselt number in Eq. (4) can be  determined only if the Reynolds number is
           known.  The Reynolds number is calculated  as
                           DG
                      Re = -
                            P
                         -                           +     Turbulent flow
                         - (0*015)(60*24) = 45,636
                            19.80 x
            The value of  L  depends on the comlations as follows:

           Dittus-Boelter correlation
           Substitution of  Eq.  (4.5-26) into Eq.  (4) gives

                          L
                          -=
                          D
                            -
                             -  (45, 636)0.2(0.707)0.6  (150  - 20)  = 58.3
                                     0.092           150 - 90
            Therefore, the required length is
                                     L = (58.3)(1.5) = 87cm

           Sieder-Tate correlation
           Substitution of  Eq. (4.5-27) into Eq. (4) gives





                                        (23.86  x  )
                     (45,636)0.2(0.707)2/3  19.80 x loq6  -0'14   150 - 20  = 49.9
                   -
                   -        0.108                          ln(150-90)
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