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AP Physics B: Ch.11 - Fluid Mechanics.
Subdivisions of matter
solids
liquids
gases
rigid
will flow
will flow
fluids
dense
and incompressible
dense
and incompressible
condensed matter
Q: what about thick liquids and soft solids?
low density
compressible
Fluid mechanics
Ordinary mechanics
Mass and force identified with objects
Fluid mechanics
Mass and force “distributed”
Density and Pressure
Density r
for element of fluid
m
r
V
mass M
volume V
m
r
V
for uniform density
mass M
volume V
units
kg m-3
Density and Pressure
Pressure p
force per unit area
F
p
A
for uniform force
F
p
A
units
N m-2
or
pascals (Pa)
Atmospheric pressure at sea level p0
on average 101.3 x103 Pa or 101.3 kPa
Gauge pressure pg
excess pressure above atmospheric
p = pg + p0
Density and Pressure
Gauge pressure pg
pressure in excess of atmospheric
gauge
atmospheric
total
p = pg + p0
typical pressures
total
gauge
atmospheric
1.0x105 Pa
0
car tire
3.5x105 Pa
2.5x105 Pa
deepest ocean
1.1x108 Pa
1.1x108 Pa
best vacuum
10-12 Pa
- 100 kPa
Example
to pump
30 cms
15 cms
The can shown has atmospheric
pressure outside.
The pump reduces the pressure inside
to 1/4 atmospheric
•
What is the gauge pressure inside?
•
What is the net force on one side?
Fluids at rest (hydrostatics)
Hydrostatic equilibrium
laws of mechanical equilibrium
pressure just above surface
is atmospheric, p0
hence, pressure
just below surface
must be same, i.e. p0
surface is in
equilibrium
Fluids at rest (hydrostatics)
Hydrostatic equilibrium
element of fluid
surface area A
height y
laws of mechanical equilibrium
S Fy =0
 pA - (p+p)A - mg = 0
-p A - rAyg = 0
(p+p)A
 p =- rgy
y
Pressure at depth h
at distance h below surface,
pressure is larger by rgh
 p = p0+rgh
pA
mg = rAyg
Question
How far below surface of water must one dive
for the pressure to increase by one atmosphere?
What is total pressure and what is the gauge pressure, at this depth?
?
Pascal’s principle
The pressure at a point in a fluid in static equilibrium depends
only on the depth of that point
Pascal’s principle
The pressure at a point in a fluid in static equilibrium depends
only on the depth of that point
Open tube manometer
(i) If h=6 cm and the fluid is mercury
(r=13600 kg m-3) find the gauge pressure in the tank
(ii) Find the absolute pressure
if p0 =101.3 kPa
Pascal’s principle
The pressure at a point in a fluid in static equilibrium depends
only on the depth of that point
Barometer
Find p0 if
h=758 mm
Pascal’s principle
The pressure at a point in a fluid in static equilibrium depends
only on the depth of that point

A change in the pressure applied to an enclosed incompressible fluid
is transmitted to every point in the fluid
Hydraulic press
p 
Fi Fo

Ai Ao
 Fo  Fi
Ao
Ai
alternative argument based on conservation of energy
Fo do  Fidi
work out = work in
Ao
di Ai  do Ao  di =
do
Ai
volume moved is same on each side
Ao
Ao
Fo do  Fi
do  Fo  Fi
Ai
Ai
Archimedes’s principle
When a body is fully or partially submerged in a fluid,
a buoyant force from the surrounding fluid acts on the body.
The force is upward and has a magnitude equal
to the weight of fluid displaced.
Fg
imagine a hole in the water-a buoyancy force exists
fill it with fluid of mass mf and equilibrium will exist
 Fb=mfg
stone more dense than water so sinks
Fb
Fg
wood less dense than water so floats
now the water displaced is less
-just enough buoyancy force
to balance the weight of the wood
Fb=Fg
Flotation
volume immersed Vi
Fb
For object of uniform density r
Fb=Fg
 rfluid Vi g= r V g
 Vi/V
= r/rfluid
Fg
Example 1
What fraction of an iceberg is submerged?
(rice for sea ice =917 kg m-3
and rsea for sea water = 1024 kg m-3)
Example 2
A “gold” statue weighs 147 N in vacuum and 139 N when immersed in salt
water of density 1024 kg m-3 . What is the density of the “gold”?
total
volume V
Fluid Dynamics
The study of fluids in motion.
turbulent
laminar
Ideal Fluid
1. Steady flow
Velocity of the fluid at any point fixed in space doesn’t change with time. This is called
“ laminar flow”, and for such flow the fluid follows “streamlines”.
2. Incompressible
We will assume the density is fixed. Accurate for liquids but not so likely for gases.
3. Inviscid
“Viscosity” is the frictional resistance to flow. Honey has high viscosity, water has small viscosity.
We will assume no viscous losses. Our approach will only be true for low viscosity fluids.
Equation of continuity
Streamlines
tube of flow
Conservation of mass in tube of flow means
mass of fluid entering A1 in time t = mass of fluid leaving A2 in time t
For incompressible fluid this means volume is also conserved.
Volume entering and leaving in time t is V
V = A1 v1 t =A2 v2 t
Therefore
A1 v1 = A2 v2
Equation of continuity
(Streamline rule)
Bernoulli’s equation
(Daniel Bernoulli, 1700-1782)
1
1
p1  2 rv12  rgy1  p2  2 rv22  rgy2
1
p  2 rv 2  rgy  constant
For special case of fluid at rest (Hydrostatics!)
p2  p1  rg(y2 - y1 )
For special case of height constant (y1=y2)
1
p1  2 r
v12
1
 p2  2 rv2 2
The pressure of a fluid decreases with increasing speed
Proof of Bernoulli’s equation
Note: same volume V with mass m
enters A1 as leaves A2 in time t
Work done at A1 in time t
(p1A1)v1 t
=p1 V
Use work energy theorem
work done by external force (pressure)
=change in KE + change in PE
W=K+ U
Work done
W  p1V - p2 V  -(p2 - p1 )V
Change in KE
1
1
1
1
K  2 mv2 2 - 2 mv12  2 rVv22 - 2 rVv12
Change in PE
U  r Vgy2 - r Vgy1
Problem
Titanic had a displacement of 43 000 tonnes.
It sank in 2.5 hours after being holed 2 m below the waterline.
Calculate the total area of the hole which sank Titanic.
Examples of Bernoulli’s relation at work
Venturi meter
Aircraft lift
Examples of Bernouilli’s relation at work
“spin bowling”