Page 65 - Essentials of physical chemistry
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Viscosity of Laminar Flow                                                    27

                                       P        dr        P


                                            r        R
                                              P    P








                                                                  d














            FIGURE 2.2 Calculus diagram of ‘‘sliding sleeves’’ of fluid flowing in a cylinder.

            Note this maintains the units described above for the sliding layers since derivative (dv=dr) has units
            of velocity=length and the total expression is a drag force. However, we need a sign reversal to
            account for the fact that the velocity decreases as the radius increases. (Let us use d ¼ l to avoid
            conflict with d=dr.):


                                              dv             dv
                                     f ¼ hA       ¼ h(2prl)      :
                                             dr              dr
            What is the force driving the fluid? We know from above that it is a difference in pressure but for the
            time being let us just call P ¼ (P 1   P 2 ) and note that a pressure is a (force=area), so we need to
            multiply the pressure by the cross-sectional area of the tube:


                                                   dv
                                                              2
                                       f ¼ h(2prl)     ¼ P(pr ):
                                                   dr
            By canceling pr and splitting the derivative into separate differentials, we obtain the variation of
            velocity in terms of the radius as

                                                   P
                                                      rdr,
                                                  2hl
                                            dv ¼
                                   v     r   P          P    r
                                  ð     ð                   ð
            which can be integrated as  dv ¼    rdr ¼         rdr.
                                   0     R  2hl        2hl   R
              Here, we use v ¼ 0at r ¼ R because the velocity is zero at the outer wall of the pipe. With these
            limits, the integrals are easy to do and we get
                                               2   2
                                         P    r   R       P    2    2
                                                              (R   r ):
                                 v(r) ¼               ¼
                                        2hl     2        4hl
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