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                                          Chapter 5
                                                   Water Hydraulics, Transmission, and Appurtenances
                                      EXAMPLE 5.27 VENTURI METER ANALYSIS
                                      The deflection of the mercury in the differential gauge of a venturi meter is y = 16.5 in. (41.91 cm) shown in Fig. 5.15.
                                                                            B  d B
                                                                                               x
                                                                             A
                                                                                               z
                                                                               d A
                                                                                               y

                                                                                      L
                                                                                             R
                                      Figure 5.15 Venturi meter analysis for Example 5.27.

                                          Determine the water flow through the venturi meter if no energy is lost between A and B, and the following are known: d =
                                                                                                                                  A
                                      12 in. (30.48 cm); d = 6 in. (15.24 cm); x = 36 in. (91.44 cm); y = 16.5 in. (41.91 cm); and z = unknown.
                                                    B
                                      Solution 1 (US Customary System):
                                      Apply the Bernoulli equation from A to B and use A as the datum:
                                                                       P    v 2      P    v 2  (  )
                                                                        A    A        B    B    36
                                                                          +    + 0 =    +    +     .
                                                                            2g            2g    12
                                                                           P A  P B  v 2 B  v 2 A
                                                                              −    =    −    + 3.
                                                                                     2g   2g
                                          From continuity equation,
                                                                        Q = A v = A v .
                                                                            A A    B B
                                                                                   2             2
                                                                         (  ∕4)(12∕12) v = (  ∕4)(6∕12) v .
                                                                                     A             B
                                                                           4v = v .
                                                                             A   B
                                                                           v = 0.25 v .
                                                                                    B
                                                                            A
                                          Apply the Bernoulli equation for L to R (s.g. mercury, Hg = 13.6):
                                                                   P A     16.5   P B  36      16.5
                                                                      + z +    =    +    + z +     × 13.6.
                                                                            12         12       12
                                                                          P     P
                                                                           A     B
                                                                             −     = 20.3 ft of water.
                                                                                  
                                                                                v 2  (0.25 v ) 2
                                                                          20.3 =  B  −    B  + 3.
                                                                                2g      2g
                                          Since g = 32.2 ft/s 2
                                                                               v = 34.49 ft∕s.
                                                                                B
                                                                                   2                3
                                                                    Q = A v = (6∕12) (  ∕4)(34.49) = 6.77 ft ∕s.
                                                                         B B
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