2.25 Advanced Fluid Mechanics
                        Solution to Prob 1, Quiz 1, Fall 2003
                        Ain S.
                        This device is a very clever, simple design that yields information about exhalation
                        pressure and volume flow rate with one single measurement. Blowing into the tube for a
                        few seconds raises the interior pressure p and moves the piston outward, which raises
                        the pressure (because of spring extension) and also the volume flow rate also (the open
                        slit area increases), until the piston reaches an equilibrium displacement  x.
                        Let the pressure inside the tube be  p, and the inlet velocity beV =Q/πR2.  Since the
                                                                                                                                             1
                        outflow area wx at the slit is very small compared with the inflow area  πR2, the velocity
                        V  at the mouth (and thus everywhere inside the tube) is negligible compared with the
                           1
                        velocityV  at the slit. Applying Bernoulli’s equation from the tube to a point in the jet from
                                        2                                                                                   
                        the slit, just outside the slit, where the pressure is atmospheric ( p=0), gives
                                                                                                                                      
                                                                                    V = 2p ρ                                                                           (1)
                                                                                     2
                        Force balance on the piston, which is essentially static during the measurement of x,
                        implies that
                                                                     
                                                                                                 pπR2 = kx                                                             (2)
                        or
                                                                                                 p= kx                                                                 (3)
                                                                                                       πR2
                                                                                 
                        Mass conservation further implies that
                                                                                                                     2p
                                                                                                Q=Vwx=                     wx                                          (4)
                                                                                                         2             ρ
                        or, inserting (3),
                                                                                                       1
                                                                                                       2
                                                                                          ⎛   2k ⎞           3 2
                                                                                    Q=⎜ ρπR2⎟  wx                                                                      (5)
                                                                                           ⎝         ⎠ 
                        Equations (5) and (3) give Q and p as functions of the piston displacement x.
                                                                     
                        SOME COMMENT
              1. Despite numerous questions about whether this Quiz would include the momentum
              theorem,  all answered in the negative both verbally and via e-mail, a large fraction of 
              the class went directly to the momentum theorem no one successfully.  This problem 
              does not lend itself easily to being solved by via the momentum theorem. (Ask at the 
              tutorials.) 
              2. Whenever you do use the CV approach, indicate clearly your CS and (in the case of 
              the momentum theorem) your frame of reference and the forces that are exerted on 
              your CV by the rest of the universe. 
              2. PLEASE WRITE LEGIBLY, differentiating clearly between p and 
, etc. 
              PROBLEM II 
                                               1 
                                                  V                    g 
                                             2             H 
                                                                U 
                                      y 
                                            D 
              The flow is inviscid and incompressible.  Since H>>D, we make the approximations that 
              it is one dimensional. 
              The vertical fluid velocity at any section is  V = Vy, t , while the horizontal gate velocity 
                                                              (   ) 
              is U = 
D . 
                     
t 
                                                 HWD = Const 
              The total fluid volume is constant,              , and hence HD = H D .
                                                                                  o  o 
                                                                                                                                                               
                                                                                                  y                                    V              U
                                                                                                                                                               
                                                                                                                                              =
                  y 
                            From mass conservation, V =
U                                             .  Furthermore,                                              . 
                                                                                                 D                                      t           t  D 
                            Apply Bernoulli’s equation between the free surface and an arbitrary pointy in the fluid, 1 
                             
 
                                  2 
                              2      
                          2           2
                                 V         
     p        V               
                                                                          
                                       •ds +            +         + gh  = 0 
                                  t              
          2             
                              1                                           	
                                                                            1 
                            Note that  
                    
               , and using the relations above, 
                                                   V • ds = Vdy 
                                U/D  y                                                2
                                 (         )                  p 
 pa          1  U              2         2 
                                                  ydy +                   +            	    y  
 H  + gy
  H = 0, after integration:
                                                                                      	 (                )        (            ) 
                                    t                                       2 
                                              H                                  D
 
                                                             2
                                                                                                                 
                               p 
 pa             1  U               2        2                             U  1                2        2 
                                            =                H  
 y  + gH
  y +                                        H  
 y  . 
                                                         (                 )        (            )                   (                ) 
                                          	     2  D 	                                                  t  D 	 
                                                                                                                        2
                            The force on the gate is the integral of the pressure difference: 
                                         H                              2                                     3                           2 
                                                                                                                                       
                               F             pdy  1  U                       1  U                 3      H                  2      H 
                                                                                           	                                            
                                     =                 =              +                   H  
                  + gH
                      , yielding: 
                                                                                                                                           
                             W                             2  D 
           2 t  D 
 	                   3                         2
                                          o                                                     
                                          
 
                                                                                              
 
                                                      2                  
                                          1                               2             g 
                               F               U             
 U                   3              2
                                      =               +                     H  +  H . 
                                                                     
                                                                         
                              PW  2  D                     
t  D          3             2 
                                                                         
                                                                                                                     F = Ma 
                            Now the equation of motion on the gate is                                                               .  Substituting: 
                                        2             3          2                       2               2     2           3      3         2                        2               2     2
                                                                                                  g                                                                             g
                               M   D             H   1  D                 1  D                     H D               H D  1  D                       1  D                    H D 
                                              =                         
                    +           o     o   =       o     o                  
                     +           o    o 
                                          2                        2          2                           2                   3              2          2                            2
                              W t                 3      D t             D  t 	               2       D                3D           D t             D  t 	                2       D
                                                                                            	                                                                            	 
                            and with further reduction we obtain the following differential equation: 
                                   3     3                         2                3     3            2 
                                                                                                       
                               H D                    M          5   D          H D D                        g        2     2    3
                                  o     o   D 
             D              
       o     o           
         H D D = 0 .
                                                                        2                                          o     o
                                    3               W           	 t                3         t 	          2