Pascal's Principle Archimede's Principle Fluid Flows Continuity - PDF - PDF

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					     Physics 111
                               Thursday,
                            November 11, 2004

• Ch 15:   Fluids
                 Pascal’s Principle
                 Archimede’s Principle
                 Fluid Flows
                 Continuity Equation
                 Bernoulli’s Equation
                 Toricelli’s Theorem
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                Announcements               111
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    • Wednesday, 8 - 9 pm in NSC 118/119
    • Sunday, 6:30 - 8 pm in CCLIR 468
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        This week’s lab will be another physics
        workshop - on fluids this time. No quiz
        this week. Bring PHYSLETS book & CD.
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                  Friday 3:20 pm in Neils 224
                  Free cookies in Neils 231 at 3 pm

                     5 class participation points

    Tom Morris on forensic accident reconstruction.
    See how the physics you’ve learned this
    semester contributes to determining how
    accidents occur and who’s responsible.
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                       Announcements                111
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        Thursday, Nov. 18, 2004
          11:50 am – 1:05 pm
        Ch. 7, 8, 9.5 - 9.6, 10 - 13

  • Hint: Be able to do the homework
  (graded AND recommended) and
  you’ll do fine on the exam!
  • You may bring one 3”X5” index card (hand-
  written on both sides), a pencil or pen, and a
  scientific calculator with you.
  • I will put any constants and mathematical
  formulas that you might need on a single page
  attached to the back of the exam.
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                       Announcements                 111
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        Thursday, Nov. 18, 2004
          11:50 am – 1:05 pm
        Ch. 7, 8, 9.5 - 9.6, 10 - 13
  Format:
  • Three sections (like homework).
    Do 2 of 3.
    Topics: Energy, Collisions, SHM, Rotation
  • One essay question. Required.
    Topics: Work, Energy
  • One section of multiple choice. Required.
    Topics: Gravity/Orbits, Torque/Rotation, SHM,
            Collisions, Energy/Work
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        We found last time that in
        hydrostatic equilibrium…


         P2 = P + ρ g(Δh)
               1


          ΔP = ρg ( Δh)
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    NOTE: our derivation here assumes a uniform
    density of molecules at a given layer in the
    atmosphere. In the real atmosphere, density
    Decreases with altitude. Nevertheless, our
    pressure and force balance diagram applies
    so long as our layer is sufficiently thin so that
    within it, the density is approximately constant.
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  If the atmosphere is in equilibrium (which would
  imply a uniform temperature and no winds
  blowing), the pressure at a given height above
  the surface would be the same around the Earth.

        The same arguments can be made for
        pressure under water. All other things being
        equal, the pressure at a given depth below
        the surface is the same.
Thurs
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                    Ch 15: Fluid Mechanics                       111
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             A scuba diver explores a reef 10 m below the
             surface. The density of water is 1 g/cc. What is
             the external water pressure on the diver?

                                           Worksheet #1
        We know the pressure at the surface of the water is
        1 atm = 101.3 kPa. The change in pressure as the
        diver drops under a 10 m column of water is given by

        ΔP = ρ g(Δh) = (10 kg/m )(9.8 m/s )(10 m)
                              3        3             2



          ΔP = 9.8 × 10 kg/m/s = 9.8 × 10 N/m
                         4         2             4       2



        P = P0 + ΔP = (101.3 + 98) kPa = 199.3 kPa
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    In solving the last problem, we applied a
    principle that we haven’t even defined yet, but
    that probably made good sense to us.

    We said that the surface pressure at the
    bottom of the atmosphere equaled the
    pressure in the surface layer of water.

    If this weren’t true, the water would fly out of the
    oceans or sink rapidly toward the ocean floor!
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        In fact, Pascal’s Principle guarantees this will
        be true. It states:
        The pressure applied to an enclosed liquid is
        transmitted undiminished to every point in
        the fluid and to the walls of the container.

        Which means, that the pressure below the
        surface of the water is equal to the surface
        pressure + the pressure due to the column
        of water above a given level.
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                           Worksheet #2
   ≥
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                                               Worksheet #2

   A container is filled with oil and fitted on both ends with
   pistons. The area of the left piston is 10 mm2; that of the right
   piston 10,000 mm2.What force must be exerted on the left
   piston to keep the 10,000-N car on the right at the same height?


       1. 10 N
       2. 100 N
       3. 10,000 N
       4. 106 N
       5. 108 N
       6. insufficient information
PI, Mazur (1997)
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                 Ch 15: Fluid Mechanics                111
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         What happens to a cork when we try to
         submerge it in water?

          It shoots right back up to the surface.


    What’s responsible for the motion of the cork?

        There must be a force acting upward on
        the cork greater in magnitude than gravity.
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                Ch 15: Fluid Mechanics             111
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        But what happens to the
        cork when it gets to the     It floats!
        surface?

          So what must be the net force on the
          cork as it’s floating on the surface?


             ZERO!         What’s changed?
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        What have we noticed about our strange
        underwater force on the cork?

        • It’s greater than gravity when the cork
          is completely submerged.

        • It’s equal to gravity when the cork floats
          on the surface, only partially submerged.
        • Our new force relates to the volume of
          the cork that’s underwater!
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                 Ch 15: Fluid Mechanics                   111
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        Archimedes had this whole process figured
        out some 2000 years ago! He said,

        A body wholly or partially submerged in
        a fluid is buoyed up by a force equal to the
        weight of the displaced fluid.

        So, the cork naturally float with just the
        right portion of its volume under the water’s
        surface so that the buoyant force upward
        from the water equals the gravitational force.
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              Ch 15: Fluid Mechanics                  111
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            If we have a cork with density of
            0.8 g/cc, what fraction of its volume
            will be below the surface in a pool of
            water when it reaches equilibrium?
 Let’s look at the free    Worksheet #3
 body diagram for                         B
 our cork.
 The gravitational force:

  W = mg = ( ρcorkVtot )g                 W

  W = (7840
                  kg
                        )Vtot   B =W
                 m2s2
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                    Ch 15: Fluid Mechanics               111
.11.

               If we have a cork with density of
               0.8 g/cc, what fraction of its volume
               will be below the surface in a pool of
               water when it reaches equilibrium?

 The buoyant force is given by
                                             B
 the weight of the displaced water.

B = mH 2 O g = ( ρ H 2OVsub )g
                      kg                  W
        B = (9800           )Vsub
                     m2s2
   Now, set this equal to           B =W
   the gravitational force...
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                                                       Phys
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                   Ch 15: Fluid Mechanics               111
.11.

              If we have a cork with density of
              0.8 g/cc, what fraction of its volume
              will be below the surface in a pool of
              water when it reaches equilibrium?
                   kg
 W = (7840                )Vtot              B
                  m2s2
                    kg
        = (9800           )Vsub = B
                   m2s2

                   kg                     W     le!
Vsub (7840             ) V                   kab
    =             m2s2    sub
                              = 0.8        ar
                                         em
                                      B =W
                   kg
Vtot (9800             ) V              R
                   2 2
                  m s        tot
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                           Worksheet #4
   ≥
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                                                      Worksheet #4


       A lead weight is fastened on top of a large solid piece of
       Styrofoam that floats in a container of water. Because of the
       weight of the lead, the water line is flush with the top surface
       of the Styrofoam. If the piece of Styrofoam is turned upside
       down so that the weight is now suspended underneath it,


       1. the arrangement sinks.
       2. the water line is below the top surface of the Styrofoam.
       3. the water line is still flush with the top surface of the Styrofoam.


PI, Mazur (1997)
Thurs
                                                     Phys
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               Ch 15: Fluid Mechanics                 111
.11.




   We’re now going to examine the behavior of
   liquids as they flow or move through pipes, the
   atmosphere, the ocean,...

   Let’s trace out the motion of a given piece or
   parcel of water as if flows through a channel.

                      These lines, which tell us
                      where a parcel has been and
                      in which direction it is going,
                      are called trajectories.
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                         If the flow is in a condition
                         known as steady state (not
                         varying) then the trajectories
                         are the same as the
                         streamlines.

        The streamlines tell us the instantaneous
        direction of motion of a parcel in a flow,
        whereas the trajectories trace out exactly
        where the parcel has actually been.
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                    Ch 15: Fluid Mechanics                     111
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   Real flows often result in turbulence -- a
   condition in which the flow becomes irregular.




                                           turbulent region
  Real flows are also often viscous.
  Viscosity describes the internal
  “friction” of a fluid, or how well one
  layer of fluid slips past another.
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                  Ch 15: Fluid Mechanics                 111
.11.




        To simplify our problems, we’re going to
        study the behavior of a class of fluids known
        as “ideal” fluids.

   1) The fluid is nonviscous (no internal friction)
   2) The fluid is incompressible (constant density)
   3) The fluid motion is steady (velocity, density
      and pressure at each point remain constant)
   4) The fluid moves without turbulence.
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        This is really just a conservation of mass
        argument. It says that if I put in 5 g of water
        each second at the left end of my hose, then
        under steady-state flow conditions, I must
        get out 5 g of water each second at the right
        end of the hose.

                                          Δmin Δmout
 5 g/s                            5 g/s       =
                                           Δt   Δt
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        For ideal fluids in steady-state (unchanging)
        flows, this must be true regardless of the
        shape of the hose. For instance, I could have
        a hose that’s narrower at the left end where
        the fluid enters the hose than it is at the right
        end where fluid leaves.

        5 g/s                                     5 g/s
   Nevertheless, the mass entering at the left each
   second must equal the mass exiting at the right.
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         v1    A1                             A2   v2
        The mass / time entering at the left side is
              Δmin    ΔV1    A1Δx1
                   =ρ     =ρ       = ρ A1v1
               Δt     Δt      Δt
        And similarly, the mass / time leaving at right is
              Δmout    ΔV2    A2 Δx2
                    =ρ     =ρ        = ρ A2 v2
               Δt       Δt     Δt
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        v1   A1                            A2   v2


        These two quantities must be equal, leaving
        us with the relationship


         ρ A1v1 = ρ A2v2       A1v1 = A2 v2
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        An ideal fluid flows through a pipe
        of cross-sectional area A. Suddenly,
        the pipe narrows to half it’s original
        width. What is the ratio of the final
        to the initial speed of the fluid flow?

        1)   4:1           Worksheet #5
        2)   2:1       2 Ar∝
        3)   1:1
        4)   1:2
        5)   1:4
                        A1v1 = A2v2
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        In examining flows through pipes in the
        Earth’s gravitational field, Bernoulli found
        a relationship between the pressure in the
        fluid, the speed of the fluid, and the height
        off the ground of the fluid.

   The sum of the pressure (P), the kinetic energy
   per unit volume (0.5ρv2), and the potential
   energy per unit volume (ρgy) has the same
   value at all points along a streamline.
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              1m 2 m
           P+    v + gy = constant
              2V    V
                1 2
             P + ρv + ρgy = constant
                2

        We can derive Bernoulli’s Equation using
        a conservation of energy argument.
                                                   Skip
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                                                Δx2
                                                       A2 P2
                                                        v2
                                          Δx1
 The change in kinetic            P1 A1               h2   g
 energy of the fluid
                                     v1    h1
 between the two ends
 must equal the net                The hatched regions have
 work done on the fluid.           the same mass of fluid.
             1        1
  ΔK = mv − mv
             2
                  2
                  2   2
                           2
                           1      Wg = −ΔU
        And the work done by          = −(mgh2 − mgh1 )
        gravity on the fluid is
        given by                      = mg(h1 − h2 )
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                                          Δx2
                                                A2 P2
                                                 v2
                                  Δx1
 Next, we consider the   P1 A1                h2 g
 work done by the
                            v1   h1
 pressure forces at each
 end of the pipe:          The hatched regions have
 W1 = P A1Δx1 = P ΔV       the same mass of fluid.
       1         1

  W2 = −P2 A2 Δx2 = − P2 ΔV

  Giving a net     Wnet = W1 + W2 + Wg
  work on
  the fluid of       = P ΔV − P2 ΔV + mg(h1 − h2 )
                        1
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                                                    Δx2
                                                              A2 P2
                                                               v2
        Now, putting it all               Δx1
                                  P1 A1                   h2      g
        together, we have
                                     v1    h1
          Wnet = ΔK
                                                1
        P ΔV − P2 ΔV + mg(h1 − h2 ) = m(v − v )
         1                                      2
                                                     2
                                                     2
                                                              2
                                                              1

                    Rearranging we get…
                        1                                 1
    P ΔV + mgh1 + mv = P2 ΔV + mgh2 + mv
     1                  2
                              2
                              1                           2
                                                                  2
                                                                  2
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             Ch 15: Fluid Mechanics                         111
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                                              Δx2
                                                        A2 P2
                                                         v2
                                        Δx1
                                P1 A1               h2    g
  Dividing by ΔV…                  v1    h1

                        2                           2
           mgh1 mv          mgh2 mv
        P+     +     = P2 +
                        1
                                +                   2
        1
           ΔV    2ΔV        ΔV    2ΔV
             Finally, identifying density…

        P1 + ρgh1 + ρv = P2 + ρgh2 + ρv
                    1
                    2
                            2
                            1
                                              1
                                              2
                                                    2
                                                    2
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   Airplane wings
        Figures copyrighted by ALLSTAR: www.allstar.fiu.edu/aero




  Bernoulli’s Principle:            P + ρv = P2 + ρv
                                         1    2         1   2
                                     1   2    1         2   2

        If v1 > v2, then P1 < P2. That creates a pressure
        gradient force known as “lift.”
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                        Ch 15: Fluid Mechanics                              111
.11.

   Curveballs
        Figure copyrighted by Cislunar Aerospace:
        muttley.ucdavis.edu/Book/Sports/instructor/curveball-01.html
                                P2
              v2                              Bernoulli’s Principle:

                                              P + ρv = P2 + ρv
                                               1
                                                     1
                                                     2
                                                          2
                                                          1
                                                                       1
                                                                       2
                                                                           2
                                                                           2

              v1
                                P1

        If v1 > v2, then P1 < P2. That creates a pressure
        gradient force that causes the “curveball.”
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           Ch 15: Fluid Mechanics                        111
.11.

        A large water tower is drained by a pipe of
        cross section A through a valve a distance
        15 m below the surface of the water in the
        tower. If the velocity of the fluid in the bottom
        pipe is 16 m/s and the pressure at the surface
        of the water is 1 atm, what is the pressure of
        the fluid in the pipe at the bottom? Assume
        that the cross-sectional area of the tank is
        much bigger than that of the drain pipe.

                       Worksheet #6

                    1 2
                 P + ρv + ρ gy = constant
                    2
Thurs
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                       Ch 15: Fluid Mechanics               111
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 Pictorial Representation:
                                        Knowns:
                                        vtop = 0
        vtop, Ptop                     Ptop = 101.5 kPa
                         g
                                        vpipe = 16 m/s
                                        H = 15.0 m
                     A          y
        H




                                        ρ = 1000 kg/m3
                                         
                                    x   g = 9.81 sm
                                                  2

                 vpipe, Ppipe
                                        Unknowns:
                                          Ppipe
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                            Ch 15: Fluid Mechanics              111
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        v = 0 m/s             Mathematical Representation:

                                Let’s look at the conditions
  15 m




                        A       at the top of the tower:
                                       1 2
           v = 16 m/s            Ptop + ρvtop + ρ gytop =
                                       2
                1 3 kg
 101.3 × 10 Pa + (10 3 )(0 m/s) 2 + (103 kg )(9.8 m/s2 )(15 m) =
                    3

                2    m                   m3


                                 = 248 kPa
            This must match the conditions for the pipe
            at the bottom of the tower...
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                        Ch 15: Fluid Mechanics                     111
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        v = 0 m/s         Mathematical Representation:

                            Let’s look at the conditions in the
  15 m




                    A       pipe at the bottom of the tower:
                                    1 2
           v = 16 m/s        Ppipe + ρv pipe + ρ gy pipe =
                                    2
                   1 3 kg                3 kg
            Ppipe + (10 3 )(16 m/s) + (10 3 )(9.8 m/s2 )(0 m) =
                                   2

                   2    m                  m


                         Ppipe + 128 kPa = 248 kPa

                              Ppipe = 120 kPa