Page 72 - Bird R.B. Transport phenomena
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§2.5  Flow of Two Adjacent Immiscible Fluids  57

                 We  may  immediately  make  use  of  one  of  the boundary  conditions—namely,  that  the
                 momentum flux r xz  is continuous through the fluid-fluid interface:

                 B.C. 1:  atx  = 0,  4  = 4                                           (2.5-4)
                 This  tells  us  that C\ =  C"; hence we  drop the superscript  and  call  both integration con-
                 stants Q.
                     When  Newton's law  of viscosity is substituted  into Eqs. 2.5-2 and 2.5-3, we  get
                                                                                      (2.5-5)


                                                         p L    C,                    (2.5-6)
                                               dx   \  L
                 These two equations can be integrated  to give
                                          ,    (Р0-Р1У
                                                                                      (2.5-7)

                                                           C l  r 1                   (2.5-8)
                                                 о  Иг      II  X
                                                 2/LL L    /л
                                                    l
                 The  three integration constants can be determined  from  the following  no-slip  boundary
                 conditions:
                 B.C. 2:                     at x  = 0,   v\ = v\ l                   (2.5-9)
                 B.C. 3:                     at x  = -b,  v\ = 0                     (2.5-10)
                                                           1
                 B.C. 4:                     atx=+b,      Ы = 0                      (2.5-11)
                 When  these three boundary conditions are applied, we  get three simultaneous  equations
                 for  the integration constants:
                 from  B.C. 2:           C\ = C\ l                                   (2.5-12)

                 from  B.C. 3:            0 =  -                                     (2.5-13)
                                                 2/x'L
                 from  B.C. 4:            0 =  -                                     (2.5-14)
                                                           M 11
                 From these three equations we get
                                              (Po ~ Pdb /V  -  д 11
                                        C, =  -                                      (2.5-15)
                                                 2L     \p
                                                      2
                                              (Po-Pi)b (  2fi ]
                                        C\=  +                  = cv                 (2.5-16)
                 The  resulting  momentum-flux  and velocity  profiles  are


                                                                                     (2.5-17)




                                                                                     (2.5-18)


                                   _ ft, - ^)b 2 Г/  у  \  / ' -                     (2.5-19)
                                                           M
                                                      11
                                       V'L    L V  + M /  V  +
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