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Design of Flow Systems                                        431

            velocity  rapidly  increases.  Simultaneously,  the  pressure  drops  rapidly,  as  illus-
            trated in Figure  8.12.  A further  increase in velocity and decrease in pressure may
            occur because of the formation  of a vena contracta which is the contraction of the
            jet  flowing  from  the orifice,  as illustrated in Figure 8.11. For liquids, if the pres-
            sure  reaches  the  vapor pressure  of the  liquid,  vaporization  will  occur.  After  the
            vena contracta, the pressure increases and the  fluid  velocity decreases, because  of
            an increase  in the  cross-sectional  area  of the valve.  The pressure  at the  outlet  of
            the valve will not reach its value at the inlet because there is a pressure loss caused
            by friction.  If the pressure rise is rapid, any vapor bubbles formed  in the valve will
            collapse instantaneously releasing large amounts of energy in a small area, which
            may be sufficient  to dent the metal.  This phenomena is called cavitation, i.e., cavi-
            tation  is the  formation  of vapor bubbles  followed  by their  sudden  collapse.  Dis-
            solved gases will also cause cavitation, such as air dissolved in water.
                 The problem that we must consider next is to relate the pressure drop across
            the  valve  to  flow  rate  and  valve  size.  After  applying Bernoulli's  equation,  Equa-
            tion 8.2, across the valve we obtain, for an incompressible fluid,

               (Pl~P2)
            E = ————                                                     (8.3)
                 P

            because the change in kinetic energy and potential energy is small, and the work
            done is zero.
                 The friction loss term, E, is given by the empirical expression
                  vo 2
            E = K——                                                      (8.4)
                 2g c
            where,  K,  an  experimentally  determined  factor,  is  the  friction-loss  factor  for  a
            valve.
                Combining Equations 8.3 and 8.4 to eliminate E and solving for, the fluid ve-
            locity in the valve orifice, v 0, we find that
                 r2 (p - ) v    /2
                   gc
                        1 P2
            v 0 =  I  ———————  I                                         (8.5)
                 I     Kp     )
                 Multiply Equation 8.5 by A  to obtain the volumetric flow rate through the
                                       o
            valve, and let p = pil, where r|  is the specific  gravity of the  fluid.  If a valve
                             w
            coefficient,  Cy, is defined by  2
                                (2 gc  r
            C v =  7.48 (12) (60) AO |  —— |                             (8.6)





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