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FLUID MECHANICS


                Example 4.10


                       A  horizontal  venturi  meter  measures  the  flow  of  oil  of  specific  gravity  0.9  in  a  75  mm

                       diameter  pipe  line.  If  the  difference  of  pressure  between  the  full  bore  and  the  throat
                                            2
                       tapping is 34.5 kN/m  and the area ratio, m is 4, calculate the rate of flow, assuming a
                       coefficient of discharge is 0.97.


                       Solution to Example 4.10


                              From Equation (3),


                                                                    2gH
                                                    Q      = c  A
                                                              d
                                                                1
                                                      actual
                                                                    m 2  − 1
                              The difference of pressure head, H must be expressed in terms of the liquid
                              following through the meter,


                                          p
                                     H  =                                        m  = 4, Cd = 0.97
                                          


                                            34  5 .  10 3
                                          =
                                            9 . 0   . 9  81 10 3


                                           . 3  142 ( 075.0  ) 2
                                      A1 =              =  . 0  00441 m
                                                                    2
                                                4

                                         =  92.3  m of  oil






                              So,


                                                                               2  . 9  81  . 3  92
                                     Actual discharge, Qactual =  97.0    . 0  00441
                                                                                   16 − 1


                                               Qactual =  0106.0  m 3  s /













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