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P. 99

7.What's   phenomenon internal friction  or viscosity ?
                  8. What does    the Newton's law of viscosity state ?
                  9.What's  unit of viscosity?
                  10. What does  Poiseuille's law state ?

                  11 What does      Stoke′s law  state?
                  12 What's   Reynolds number?


                                              7.10 Problems

                         Air    blows  out    through  the  horizontal  tube    AB  as  shown  in

                  fig.7.12 .The volume rateQ of air  is equal to 1 liter per minute .There
                                                                       2                 2
                  are two cross section in tube :S               2sm ,S           5 , 0 sm . Calculate  the
                                                             1
                                                                            2
                                                             difference of water levels   in a tube
                                                             which connects two parts of tube  AB.
                                                                                             kg
                                                             Air  density        a    , 1 32     3  ,  water
                                                                                                m

                                                                                      kg
                                                             density    w    1000         3  .
                                                                                         m
                                                                   Solution
                                                                   According to Bernoulli's equation

                                                               for  the  horizontally  located  tube  of
                        Figure7.12
                                                             flow  of ideal liquid or gas:


                                                             2                   2
                                                       a  v   1          a  v   2
                                               p                 p 
                                                 1                  2
                                                         2                   2
                  we will get

                                                                       2     2
                                                 p   p    p       (v    v 1  ).
                                                                        2
                                                              2
                                                        1
                                                                   2
                  Taking into account  that

                                                                          Q                Q
                                    Q    S  v    S    v       v         and   v 
                                           1    1     2    2         1                2
                                                                          S                S
                                                                           1                2
                  therefore
                                                           2     2      2
                                                       Q     S 1   S 2  
                                                     p                  .
                                                        2      S 1 2   S 2 2  
                                                             
                                                                          
                  Substitute all values  and obtain   p           750  Pa



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