# Pascal s principle Hydraulic systems Force on a piston

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```					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
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

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