From last class:
Pressure, depth and Pascal s
Fluid statics
Law
• What is a fluid?
Density !
Pressure!
• Fluid pressure and depth
Pascal s principle
• Buoyancy
Archimedes principle
In a static fluid, with uniform density ρ,
Pressure at depth h = pressure acting on surface
+ pressure due to height of liquid
Lecture 2
Dr Julia Bryant
Fluid dynamics!
• Reynolds number
• Equation of continuity!
• Bernoulli s principle
• Viscosity and Turbulent flow
• Poiseuille s equation
ph = p0 + F/A
F = weight of column liquid of cross sectional area A
F = mg
p0 pressure acting on surface
M=ρV
= ρ Ah
F/A = ρ gh
pressure3.pdf
pascal.pdf
h
ph = p0 + ρgh
A
These lecture notes: http://www.physics.usyd.edu.au/~jbryant/Fluids/
Another resource:
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http://www.physics.usyd.edu.au/teach_res/jp/fluids/wfluids.htm
!
Weight of
column
of liquid
F
Liquid – uniform density ρ
2
Example:
The pressure within a uniform stationary
fluid is the same at all points in the
same horizontal plane.
Estimate the difference in fluid pressure between the
neck and base of a bottle of wine when
(a) upright and (b) cellaring (lying down)
ρ  = 1.08 x 103 kg m-3
h
h = 0.23 m
∆p = ρ g ∆h = 1.08 x 103 x 9.8 x 0.23
The pressure exerted by a static fluid depends only upon the depth of
the fluid, the density of the fluid, and the acceleration of gravity
= 2434Pa = 2.4kPa
ph = p0 + ρ g h
Static pressure does not depend upon mass or surface area of liquid
and the shape of container due to pressure exerted by walls.
A 175cm tall person has a difference
in blood (density 1.06 x 103 kg m-3)
pressure of 18179Pa between their
head and feet. Work it out.
DEMO
3
Example:
In liquids p=p0 + ρgh but what about gases?!
Water is in a U-tube. Oil is added to
one side until it is a height d above
the water, which has risen a distance
H.
Pressure at the interface height is
Left tube: pL= ρw g H
Right tube: pR= ρoil g (H+d)
pL = pR
!
Gas pressure!
!
oil
water
ρw g H = ρoil g (H+d)
ρoil = H
ρw H+d
pV = N k T
d
What is the density of the oil?
4
ρ/ ρw is the specific gravity
Interface
poil=pwater
5
pV = n R T
piVi = pfVf !
Ti
Tf
p is the gas pressure (Pa), "
V is the volume of the gas (m3), "
T is the gas temperature (K), "
N is the number of molecules and "
n is the number of moles of the gas (mol) "
Boltzmann constant k = 1.38x10-23 J.K-1"
Universal gas constant R = 8.314 J.mol-1.K-1"
k = R / NA R = k NA
"
" Avogadro's constant NA = 6.023x1023 mol-1"
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Isothermals pV = constant
Measuring relative pressure using a DEMO
barometer or manometer p = atmospheric pressure
180
160
piVi = pfVf
Ti
Tf
pressure p (kPa)
140
120
D
D
100 K
100
200 K
Manometer
300 K
80
A
400 K
B and C are at
the same level
so
pB = p C
60
B
C
40
20
0
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
Pressure at base
p1 = pA + ρ g y 1
0.40
p2 = p 1
3
volume V (m )
Gas laws (fixed quantity of gas)
"
Boyle's Law (constant temperature) p = constant / V"
Charles Law (constant pressure)
V = constant x T
(constant volume)
p = constant x T "
Pressure at base
p2 = p 0 + ρ g y 2
p0 + ρ g y 2 = pA + ρ g y 1
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Measuring relative pressure
pA - p0 = ρ g( y2 - y1) = ρ g h
p is the absolute pressure
p - p0 is the gauge pressure
Why does a brain tumor affect the
spinal cord?
Mercury Barometer
measures atmospheric
pressure
For example patm = 760 mmHg"
P
" atm = ρgh"
ρ = 13.6 x 103 kg.m-3 "
g = 9.8 m.s-2"
h = 760 mm = 760 x 10-3 m"
Patm = (13.6 x 103)(9.8)(760 x 10-3)
Pa = 1.01 x 105 Pa"
patm"
tumor
Increased
pressure
transmitted down
spinal cord
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Pascal's Principle
8
1653 Blaise Pascal (1623 – 1662)
10
Pascal's Principle
Pressure applied to an enclosed fluid is transmitted undiminished
to every portion of the fluid and walls of the containing vessel.
force
1653 Blaise Pascal (1623 – 1662)
Pressure applied to an enclosed fluid is transmitted undiminished
to every portion of the fluid and walls of the containing vessel.
ph
ph
p0ʼ
p0
p0
2.4kPa + Pincease
(0,0)
DEMO
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h
(0,0)
h
Linear relationship between pressure and depth.
If the pressure at the surface increases then the pressure at
a depth h also increases by the same amount.
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Tennis Ball Impact on Eye
How can a person easily lift a car?
A blow to the eye by a tennis ball can cause more
damage than one might expect because of the
transmission of the pressure to the back of the eye
• The cornea on the front of the eye is tough and may feel the pain of the
impact but without damage.
• Pascal s principle means that the pressure is transmitted through the fluid,
from the front of the eye, undiminished to all parts of the eye. In this way the
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retina can be severely damaged or torn.
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• A piston with area A1 exerts force F1 on a fluid, which connects a larger
piston of area A2.
• Based on Pascal s principle, the pressure is the same on both cylinders.
F1
P=
F2 =
F1
F1
F
= 2
A1
A2
F2
F1
A2
F >> F1
A1 1
pA1
• Small piston moves a distance h1 small force + large distance
==> large lifting force
• Volume of fluid displaced ∆V=h1 A1
over small distance
• Large piston moves h2 with ∆V=h2 A2
so
A1
F2
h2 =
h << h1
F1
A2 1
pA1
h1
oil
h1
oil
F2
F2
h2
pA2
A1
h2
pA2
A1
A2
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A2
16
BUOYANCY - FLOATING AND SINKING
F2 =
A2
F >> F1
A1 1
h2 =
Why do ice cubes float on water?
Less dense than water.
Yes, but why does something with less
density than water float?
Why does a hot air balloon rise?
A1
h << h1
A2 1
• But work done= F1 h1
=
(F2 x
A1
A2
)(h2 x
A2
A1
) = F2 h2
Energy is conserved
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If I suspend a block on a
rope, what force do we
need to pull up with on
the rope to make the
block hang steady?
Buoyancy
rope
•  When a solid object is wholly
or partly immersed in a fluid,
the fluid molecules are
continually striking the
submerged surface of the
object. "
•  The forces due to these
impacts can be combined into
a single force the buoyant
force which counteracts the
weight."
T=?
If I immerse the block in water,
what happens to the tension in
the rope?
m
W
T = W - FB
T
FB
W
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Thin sack filled with water.
Fb Weight of water mg=Fb
If Fb > Fg body floats.
Weight of object = weight of fluid displaced by object "
If Fb < Fg body sinks.
Fg Volume of displaced water = volume of object
A body floats in any
liquid with density
ρfluid > ρbody
Fg
Fb
Fg
Fb > Fg
Replace sack with stone
Fb msg>Fb ===> sinks
Fb < Fg
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Weight of object > weight of fluid displaced by object "
Volume of displaced water = volume of object
Replace sack with wood
mwg<Fb ==>floats (ρwood<ρwater)
Fb Weight of object < weight of fluid displaced by object "
of displaced water < volume of object"
Fg Volume
Weight of liquid displaced by submerged part "
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of the object = weight of object
How high will it float?
How high will it float?
Wood breaks surface and displaces less
water until
- What fraction of an iceberg is under water?
Fb = mwoodg=m'g
where m' and V' are the mass and volume of
the water displaced
less
dense
more
dense
ρV'g = ρwoodVwoodg
Vsubmerged
ρwood
Vwood =
ρ
Fb Fraction of block fluid
F submerged is ρwood / ρ
g
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Water expands on freezing
by 10%.
Density of ice is 0.9g/cm3
Fraction of iceberg
submerged is
ρice / ρ water= 0.9/1.0
Therefore 90% of the
iceberg is submerged.
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Example problem:
Another example problem:
The pressure on the surface of a lake is atmospheric pressure, Pat.
(a)  At what depth is the pressure twice atmospheric pressure?
(b) If the lake was full of mercury, at what depth is the pressure 2Pat?
Blood flows into the aorta through a circular opening of
radius 0.9cm.
If the blood pressure is 120 torr, how much force must be
exerted by the heart?
(a) p = pat + ρgh
When p=2pat,
2pat = pat + ρgh
1 torr = 133.322Pa = 133.322 N.m-2
pat = ρgh
120 torr =120 x 133.322 N.m-2
h = pat /ρg
F = pA
= 1.01 x 105/(1x 103 x 9.8) = 10.3m
F = 120 x 133.322 x π x (9x10-3 )2 N
(b) hHg = pat /ρHgg
= 4.07 N
= 1.01 x 105/(13645 x 9.8) = 75cm
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And another example problem:
A 1500kg car rests on four tires, each of which is inflated to a gauge pressure
of 200kPa. What is the area of contact of each tire with the road if the four
tires support the weight equally?
Why do your ears feel different when you dive into deep water?
Why does atmospheric pressure change as you go up a mountain?
F = pgaugeA
A=F/pgauge
Fcar=1500 x 9.8
A = 1500 x 9.8/(4 x 2 x 105 ) = 1.8 x 10-2 m2 =184 cm2
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