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116  FLOW  OF  FLUIDS



                  EXAMPLE 6.14                                                  4(39.11)
                                                                                             - 157,100
                  Pressure Drop and Void Fraction in Liquid-Gas  Flow   Re =  n(32.2)(0.2557)3.85(10-5) -
               A mixture of  an  oil and hydrogen at  500psia  and 200°F enters a
               3 in. Schedule 40 steel line. Data are:                 f= 0.0202,
                  Oil:  140,000 Ib/hr, 51.85 Ib/cuft, 2700 cfh, viscosity 15 cP.
                  Hydrogen:  800 lb/hr,   0.142 lb/cuft,   5619 cfh,   viscosity
               2.5(10-7) Ibf sec/sqft.                             compared with 53.0 by the LMC method.
                                                                      Voidfraction by Eq. (6.104):
               The  pressure  drop  in  1OOft  of  line  will  be  found,  and  also  the
                                                                                           =
               voidage at the inlet condition.                        E~  = 1 - 1/GL = 1 - l/m 0.413,

                   Re,=--   4k   -   4(140,000/3600)               compared with input flow condition of
                       nDg,,u   n(0.2557)(32.2)0.15  ’
                             4(800/3600)                              E=--   QG     5619   =0.675.
                  Re  -                  7  = 137,500,                   Q, + QL - 5619 + 2700
                      -  n(0.2557)(32.2)(2.5)(10-  )
                     E
                    - = 0.00059.                                      Method of  Premoli [Eqs. (6.105) and (6.106)]:
                    D
                                                                      Surface tension u = 20 dyn/cm, 0.00137 lbf/ft,
                   Round equations:
                                 1.6434        0.0272,  liquid,
                        f=[ln(0.135~/D  + 6.5/ReI2= { 0.0204,  gas,
                                                                         -  -      16(38.89)’      = 64,118,
                                                                           n2(32.2)(0.2557)’(51.85)(0.00137)
                         = 18.27 psf/ft,                               Re = 19,196,
                                                                       E, = 1.578(19196)-0~’9(51.85/0.142)0-22 0.8872,
                                                                                                   =
                                                                       E, = 0.0273(6411.8)(19196)-0~51(51.85/0.142)-0~08 7.140,
                                                                                                          =
                       Xz = 18.27/0.1633 = 111.8.                       y = 5619/2700 = 2.081,
                  Lockhart-Martinelli-Chisholm:                       YE, = 2.081(7.140) = 14.86.
                  c = 20 for TT regime (Table 64,                  Clearly, this term must be less than unity if  Eq. (6.105a) for S is to
                         c1                                        be  valid,  so  that  equation  is  not  applicable to this  problem  as it
                         x xz-
                   &  1 + - + - - 2.90,                            stands. If  YE, is replaced by y/E, = 0.2914,  then
                   :.  (APIL) two phase = &(AP/L),  = Z.gO(18.27)
                                                                                  2.081
                                                                                               =
                                  = 53.0psf/ft,  36.8 psi/lOO ft.     S = 1 + 0.8872(~ 0.2914)~.~ 2.02,
                                                                                      -
               Check with the homogeneous model:                   and the voidage is
                          8oo                                                 5619
                    X=  140,000 -I- 800  = 0.0057 wt  fraction gas,   E = 5619 + 2.02(2700) = 0.51’
                    ’=  [m 3.13(1OW4) 1             sqft  ’        which is a plausible result. However, Eqs. (6.105) and (6.105a) are
                        0.0057
                              +  0.9943
                                        -1=3.85(10-5)-  lbf sec
                                                                   quoted  correctly from  the  original paper;  no  numerical examples
                                                                   are given there.




               Hewitt  (1982).  A  procedure  for  stratified  flow  is  given  by   17-26  (1971); cited by Hewitt, 19821 gives the void fraction in terms
               Cheremisinoff and Davis [AZChE J. 25,  1 (1979)].   of  the incoming volumetric flow rates by the equation
                   Voidage of the holdup in the line is different from that given by
               the proportions of the incoming volumetric flows of  the two phases,   EG = Qc/(Qc + SQJ,      (6.105)
               but  is  of  course  related  to  it.  Lockhart  and  Martinelli’s  work   where S is given by the series of  equations
               indicates that the fractional gas volume is
                                                                      S = 1 + El[y/(l +YE,)  YE,]"^,
                   E = 1 - l/4L,                           (6.104)    E, = 1.578 Re-0.’9(pL/pG)0.22,          (6.105’)

               where  +L  is defined in Table 6.8.  This relation has been found to   E, = 0.0273 We Re-o.5i(p~/p~)-o.08,
               give high values. A correlation of  Premoli et al.  [Termotecnica 25,   y = QG/QL,  Re = DG/p,,   We = DG2/apL.
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