Page 97 - Numerical Methods for Chemical Engineering
P. 97

86      2 Nonlinear algebraic systems




                      1




                      1
                    viscsit in a s  1









                      1  2
                       1  −1        1            1  1         1  2
                                       sear r ate in s −1
                   Figure 2.16 Shear-dependent viscosity of poly(styrene) melt at 453 K, estimated from the Carreau–
                   Yasuda model.


                   Abdel-Khalik et al. (1974) fit the poly(styrene) melt data at 453 K of Ballenger et al. (1971)
                   to obtain
                                             4
                         η ∞ = 0  η 0 = 1.48 × 10 Pa s  λ = 1.04 s n = 0.398 a = 2    (2.95)
                   The predicted shear-thinning behavior is shown in Figure 2.16.
                     With this nonlinear dependence of the viscosity upon the local shear-rate, numerical
                   solution of the equation of motion is now required. For 1-D laminar flow, the equation of
                   motion
                                      Dv      ∂
                                     ρ    = ρ  v + ρ(v · ∇ v) = ∇ · τ − ∇P            (2.96)
                                       Dt    ∂t
                   reduces to
                                            dτ yx  dP            dv x
                                         0 =     −      τ yx = η(˙γ )                 (2.97)
                                             dy    dx             dy
                   τ yx is the shear-stress and P is the dynamic pressure. For a constant dynamic pressure
                   gradient, the equation of motion

                                                dτ yx    P
                                                     =                                (2.98)
                                                 dy      x
                   yields a shear-stress that varies linearly with y,
                                                            P

                                           τ yx (y) = τ yx (0) +  y                   (2.99)
                                                             x
                   By symmetry, the velocity gradient mid-way between the plates is zero, and thus the shear-
                   stress at y = B/2 is zero. Hence,
                                           B                  P    B

                                       τ yx   = 0 = τ yx (0) +                       (2.100)
                                           2                  x    2
   92   93   94   95   96   97   98   99   100   101   102