Page 226 - Bird R.B. Transport phenomena
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210  Chapter 7  Macroscopic Balances  for  Isothermal Flow  Systems


      SOLUTION             (a)  Mass  balance.  For steady flow the mass balance gives
                                                     VL\  =  IV 2  ОГ  P]V ] S l                (7.6-1)
                           For  a fluid  of constant density, this gives
                                                              V\  \
                                                                                                 (7.6-2)
                           (b)  Momentum  balance.  The downstream component of the momentum balance is

                                                            -  v 2 w 2 )  -  p 2 S 2 )          (7.6-3)
                           The  force  F^  is  composed  of  two parts: the viscous  force  on the cylindrical  surfaces  parallel
                                      s
                           to  the direction  of flow, and the pressure  force  on the washer-shaped  surface  just  to the right
                           of  plane  1 and  perpendicular  to the flow axis.  The former  contribution we  neglect  (by intu-
                           ition) and  the latter  we  take  to be p\{S  — Sj)  by  assuming  that the pressure  on the  washer-
                                                          2
                           shaped  surface  is the same as that at plane 1. We  then get, by using  Eq. 7.6-1,
                                                                  -  v 2 )  -  p 2 S 2 )         (7.6-4)
                           Solving  for  the pressure  difference  gives
                                                                    \  -  v 2 )                  (7.6-5)
                           or,  in terms  of the downstream  velocity,

                                                                                                 (7.6-6)

                           Note that the momentum balance predicts  (correctly) a rise in pressure.
                           (c)  Angular  momentum  balance.  This balance is  not needed.  If we  take the origin  of  coor-
                           dinates  on  the  axis  of  the  system  at  the  center  of  gravity  of  the  fluid  located  between
                           planes  1 and  2, then  [t }  X  u j  and  [r 2  X u ]  are both zero, and  there are  no torques  on the
                                                             2
                           fluid  system.
                           (d)  Mechanical  energy balance. There  is  no  compressive  loss,  no  work  done  via  moving
                           parts, and no elevation  change, so that

                                                     E v  = 4v\  ~ z> + p  (P\ ~ Pi)             (7.6-7)
                                                              v
                           Insertion  of  Eq. 7.6-6  for  the pressure rise then gives, after  some rearrangement,

                                                                                                 (7.6-8)

                           which is an entry in Table 7.5-1.
                               This example has shown how to use the macroscopic balances to estimate the friction loss
                           factor  for  a simple  resistance  in a flow system.  Because  of  the assumptions  mentioned  after
                           Eq.  7.6-3, the results  in Eqs. 7.6-6 and 8 are approximate.  If great  accuracy is needed, a correc-
                           tion factor based  on experimental data should be introduced.






       EXAMPLE   7.6-2     A diagram  of  a liquid-liquid  ejector  is shown  in Fig. 7.6-2.  It is desired  to analyze  the  mixing
                           of  the two  streams, both  of  the same fluid, by  means  of  the macroscopic balances.  At  plane 1
      Performance  of  a   the  two  fluid  streams  merge.  Stream la  has  a  velocity  v  and  a  cross-sectional  area  ^S  and
                                                                                                 u
      Liquid-Liquid  Ejector  stream lb  has a velocity  \v  and a cross-sectional  area \S V  0 Plane 2 is chosen far enough down-
                                                Q
                           stream that the two  streams  have  mixed  and the velocity  is  almost uniform  at v . The flow is
                                                                                           2
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