Pascal s principle Hydraulic systems Force on a piston

Document Sample
Pascal s principle Hydraulic systems Force on a piston Powered By Docstoc
					Fluids
    Liquids and Gases




                   Chapter 11
Density

  Density is the amount of mass in 1 m3.

               mass                           kg
    density                        Units :
                                              m3
              volume
        m
          and m  V
        V

What is the mass of 5200 cm3 of 1060 kg/m3 blood?
                                          3
                kg          3  1m     
   m  V  1060 3   5200cm           5.51kg
                m               100cm 
Density
Pressure
 Pressure is the force exerted
 by a fluid on an area of 1 m2.


           F
        P
           A
                                               N
                            Unit: pascal  Pa  2
                                               m
        F  PA
Air pressure

   Standard atmospheric pressure at sea level is
       101,300 Pa = 1 atmosphere = 1 atm
Example
How much force does the air exert on the liquid in
the tube that has a 2 cm2 surface area?




F  PA
                                        2
           N        2  1m     
F  101,300 2   2cm           20.26 N
           m            100cm 
Pressure at depth in a liquid
Pressure is the same everywhere at the same depth.
Consider a block of liquid at rest.

 F  Fbottom  Ftop  Fg  0
 F  Pbottom A  Ptop A  mg  0
 Pbottom    Ptop  A  mg
                          V  g    Ah  g   hg
 Pbottom    Ptop 
                     A
                       
                                        A
 Pbottom  P    gh
             top
                                            pressure difference

                                            Pdepth   gh
Example 4 The Swimming Hole
Points A and B are located a distance of 5.5 m beneath the
surface of the water. Find the pressures at A and B.
 PA  Psurface  Pdepth  Psurface   gh
                                pressure difference for 5.5m
       atmospheric pressure
                          kg     m
 PA  101,300 Pa   1000 3  9.8 2   5.5 m 
                          m      s 
 PA  155, 200 Pa
A and B are at the same                            Psurface
depth so their pressures
are the same.
                                                     Pdepth
 PB  155, 200 Pa
Blood pressure
What is the difference in blood pressure between a person's
heart and their feet?


                        Pdepth   gh
                                          kg   m
                        Pdepth    1060 3  9.8 2  1.35m 
                                          m    s 
                        Pdepth    14, 024 Pa
Pressure



Points A, B, C, and D are all ...
   • At the same level.
   • In the same liquid .
   • At the same pressure.
Barometer
             no air pressure           A barometer is an instrument used
                                       to measure atmospheric pressure.

                                          PA  Psurface  Pdepth
                Psurface  0
                                          Patmosphere  0   gh
                      liquid
                     mercury
                                       How tall would mercury be for
              Pdepth                  101,300 Pa air pressure?

                                                           kg      m
                                       101,300 Pa  13, 600 3   9.8 2  h
                                                           m       s 
                        air pressure   h  0.760 m  760 mm


A and B are at the same level so       For water h = 10.3 m or about 34 feet.
their pressures are the same.
Absolute pressure and Gauge pressure


                     Tire pressure is measured in
                     customary units of psi.
                        [ psi means pounds/inch2 ]



 Inside the tire the gauge pressure is 30 psi above
 normal atmospheric pressure.
 Outside the tire atmospheric pressure is 15 psi.
 Inside the tire the absolute pressure is 45 psi.
       30 psi + 15 psi = 45 psi.
Absolute pressure and Gauge pressure

                  gas pressure
                                                        Psurface

   PA  Psurface  Pdepth                               Pdepth


PA is the absolute pressure at point A.
Psurface is the atmospheric pressure Patmosphere .   A and B are at
                                                     the same level so
Pdepth is how much higher the gas                   their pressures
                                                     are the same.
        pressure is above atmospheric pressure.
Pdepth is the gauge pressure.
Open-tube manometer blood-pressure gauge
Blood pressure cuff measures a
patient's blood pressure in units
of the height of mercury in a tube.

    Pdepth   gh

                             Psurface
                            Pdepth         Pair in the cuff


                                        Blood pressure is
                                        gauge pressure.
Measuring blood pressure
Blood pressure in the heart
should be measured at the same
horizontal level as the heart
(usually at the upper arm).

                  P  130 / 85
                 Pdepth  105 mm of Hg



                 P  235/190

                                      760 mm of Hg 
                 Pdepth  14,024 Pa                 105 mm of Hg
                                      101,300 Pa 
Pascal's principle

Increasing the pressure at one place
in an enclosed fluid increases the
pressure by the same amount
everywhere in the fluid.




This principle explains the operation
of all hydraulic systems.
Hydraulic systems

             Force = Pressure  Area

 Force on a piston is proportional to the piston area.


    Small piston  Small force
    Large piston  Large force
Example 7 A Car Lift
Find gauge pressure needed to raise the car.
Find force needed at the small piston.
  Car and large piston weigh 20,000 N.
  Density of the hydraulic oil is 800 kg/m3.
  Height h is 2 m. Ignore the weight of the pistons.

         Flarge       20, 000 N
  PB                         2
                                  282,900 Pa
         Alarge       0.0707m                          Small piston
                                                         r = 0.012 m
  PA  PB (same level in same liquid)                    A = 4.52 x 10-4 m2

                              Fsmall                   Large piston
  PA  Psurface  Pdepth             gh               r = 0.15 m
                              Asmall                      A = 0.0707 m2
             Fsmall             kg      m
  PA                      800 3  9.8 2   2m 
         4.52 104 m 2         m       s 
                        Fsmall
  282,900 Pa                 4
                                    15, 680 Pa
                    4.52 10 m   2

  Fsmall  120.8 N
Archimedes' principle

 The upward force that fluids
 exert on partially or fully
 submerged objects is called
 a buoyant force.

 The buoyant force is equal to
 the weight of the fluid
 displaced by the object.

      Fbuoyant   Fg displaced fluid   mdisplace fluid  g

      Fbuoyant    fluidVolumedisplaced  g  Vg
Example 9 A Swimming Raft
How deep is the raft submerged?                         Pine wood
                                                        550 kg/m3

F  0  Fg  Fbuoyant (equilibrium)
0  mraft g   waterVsubmerged g
0     raftVraft  g   waterVsubmerged g
 waterVsubmerged   raftVraft
 water  Araft hsubmerged    raft  Araft hraft 
                                    kg 
                raft hraft      550 3   0.3m 
hsubmerged                        m 
                                                    0.165m
                 water             1000 3
                                          kg
                                         m
Example 9 A Swimming Raft
Since the density of the raft was                   Pine wood
                                                    550 kg/m3
less than the density of the water,
the submerged depth was less
than the thickness of the raft.
If the density of the raft were
greater than the water, the raft
would sink because it could not
displace a weight of water equal
to its own weight.
                                      kg 
                     raft hraft  550 3   0.3m 
                                       m 
       hsubmerged                                   0.165m
                      water               kg
                                      1000 3
                                           m
Archimedes' principle
                            The fluid in a charged battery
                            has a higher density than the
                            fluid in a discharged battery.




Green ball floats when
the fluid density is more
than the ball density.
Fluid dynamics: vocabulary


 Streamlines - paths of fluid particles
 Steady flow - constant velocity
 Unsteady flow - variable velocity
 Turbulent flow - erratically variable velocity
 Compressible - variable density (example: gases)
 Incompressible - constant density (example: liquids)
 Viscous flow - does not flow easily (example: honey)
 Non-viscous flow - flows easily (example: water)
 Ideal fluid - incompressible, non-viscous fluid
    (Our focus will be on the behavior of ideal fluids.)
Mass flow rate
 Mass of fluid flowing past a point in one second is
 called the mass flow rate.
 The (blue) mass m with velocity v in section 1 will
 flow past the end of the tube in time Δt.

                 mass  Volume    Avt 
mass flow rate                              Av
                 time     t          t


                                m

                                                    kg
                             Mass Flow Rate unit:
                                                     s
Continuous flow




 A single tube with wide and narrow sections has different
 velocities in each section, but the same mass flow rate.

          mass flow rate section 1   mass flow rate section 2

          1 A1v1   2 A2 v2    continuity equation 
Continuous flow




The density is the same throughout an incompressible fluid.
            1 A1v1   2 A2 v2 (continuity equation)
                    1   2           (incompressible fluid)
Therefore         A1v1  A2v2          (volume flow rate)
 volume flow rate section 1   volume flow rate section 2
For an incompressible fluid, the volume flow rate is the same
everywhere in the tube.
Example 12 A Garden Hose
Hose fills an 8 x 10-3 m3 bucket in            cross sectional area
20 seconds.                                    of 1.6x10-4 m2

Find the volume flow rate.
Find the mass flow rate.
                                                cross sectional area
Find the velocity for each example.             of 0.8 x10-4 m2

                      volume 8 103 m3           m3
 volume flow rate Q                    4 104
                       time     20s               s
 Q  Av
              m3
              4
      4 10
   Q            s  2.5 m (wide area)
 v 
   A 1.6 104 m 2      s
              m3
      4 104
   Q            s 5m
 v         4
                          (narrow area)
   A 0.8 10 m   2
                      s
Pressure changes along a flowing fluid




                                                Pdepth




  Pressure change provides       Pressure change is due to
  the net force needed to        the difference in depth in
  accelerate the fluid.          the fluid.
  Pressure must drop to
  allow the fluid to speed up.
Bernoulli's equation for an ideal fluid
Bernoulli's equation
describes the relationships
between pressure, velocity,
and height in a flowing fluid.



            P  v  gy1  P2  v  gy2
             1
               1
               2
                    2
                    1
                                        1
                                        2
                                              2
                                              2

Pressure
 Velocity                  Based on principles of work
  Height                   and mechanical energy.
Applications for Bernoulli's equation
   P  v  gy1  P2  v  gy2
    1
            1
            2
                  2
                  1
                                              1
                                              2
                                                  2
                                                  2

 Conceptual Example 14 Tarpaulins and Bernoulli’s Equation

 When the truck is stationary, the
 tarpaulin lies flat, but it bulges outward
 when the truck is speeding down
 the highway.

 Explain.
Applications for Bernoulli's equation

     P  v  gy1  P2  v  gy2
      1
          1
          2
              2
              1
                               1
                               2
                                   2
                                   2
Applications for Bernoulli's equation

     P  v  gy1  P2  v  gy2
      1
          1
          2
              2
              1
                               1
                               2
                                   2
                                   2
Applications for Bernoulli's equation

     P  v  gy1  P2  v  gy2
      1
          1
          2
              2
              1
                               1
                               2
                                   2
                                   2
The End

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:25
posted:8/8/2012
language:English
pages:34