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FLUID MECHANICS




                Example 5.2




                                               3
                       A  pipe  carrying  0.06  m /s  suddenly  contracts  from  200  mm  to  150  mm  diameter.
                       Assuming that the vena contracta is formed in the smaller pipe, calculate the coefficient of

                       contraction if the pressure head at a point upstream of the contraction is 0.655 m greater
                       than at a point just downstream of the vena contracta.





                       Solution to Example 5.2


                       Inserting this expression for the loss of head in Bernoulli’s equation,


                                                                        2
                                                                       
                                      p 1  +  v 1 2  =  p 2  +  v 2 2  +  v 2 2    1  − 1
                                                               
                                       g   2g    g   2g   2g   C C  
                                                               
                                                                       
                                       p −  p    v  2      1   2   v  2
                                                                
                                         1   2  =  2  1  +   − 1   −  1
                                                        
                                                                
                                          g     2 g    C         2 g
                                                          C      
                                       p  − p
                              Given,      1  2  =  . 0  655
                                          g



                              Using the continuity of flow Q = Av for velocity v1 and v2.


                                          Q
                                     v =
                                      1
                                          A 1

                                           . 0  06 4
                                         =
                                           ( ) 2.0  2



                                          1=  . 91 m  s /


                                           Q
                                     v =
                                      2
                                          A 2


                                           . 0  06 4
                                         =           3=  4 .  m  s /
                                           ( 15.0  ) 2





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