subdivisions of matter
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AP Physics B: Ch.11 - Fluid Mechanics. Subdivisions of matter. solidsliquidsgases rigidwill flowwill flow dense dense low density and incompressible and incompressible compressible. fluids. condensed matter. - PowerPoint PPT PresentationTRANSCRIPT
Subdivisions of matter
solids liquids gases
rigid will flow will flow
dense dense low density and incompressible and incompressible compressible
fluids
condensed matter
Q: what about thick liquids and soft solids?
AP Physics B: Ch.11 - Fluid Mechanics.
Fluid mechanics
Ordinary mechanics
Mass and force identified with objects
Fluid mechanics
Mass and force “distributed”
Density and Pressure
Density
mV
mV
for element of fluid
mass Mvolume V
for uniform density
mass Mvolume V
units kg m-3
Density and Pressure
Pressure p
p FA
p FA
force per unit area
for uniform force
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
p = pg + p0pressure in excess of atmospheric
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
atmosphericgauge
total
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 inequilibrium
Fluids at rest (hydrostatics)
Hydrostatic equilibrium
laws of mechanical equilibrium
(p+p)A
pA
y
element of fluidsurface area A
height y
pA - (p+p)A - mg = 0
Fy =0
p A - Ayg = 0
mg = Ayg
p =- gy
p = p0+gh
at distance h below surface,
pressure is larger by gh
Pressure at depth h
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 (=13600 kg m-3) find the gauge pressure in the tank
(ii) Find the absolute pressureif 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 ifh=758 mm
Fodo Fidi
diAi do Ao di =Ao
Ai
do
Fodo Fi
Ao
Ai
do Fo Fi
Ao
Ai
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 fluidis transmitted to every point in the fluid
Hydraulic press
p Fi
Ai
Fo
Ao
Fo Fi
Ao
Ai
alternative argument based on conservation of energy
work out = work in
volume moved is same on each side
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.
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
wood less dense than water so floats
now the water displaced is less-just enough buoyancy forceto balance the weight of the wood
Fb=Fg
Fb
Fg
Fg
Example 1
What fraction of an iceberg is submerged?(ice for sea ice =917 kg m-3 and sea for sea water = 1024 kg m-3)
Fb
Fg
Fb=Fg
fluid Vi g= V g
Vi/V = /fluid
volume immersed Vi totalvolume V
Example 2
A “gold” statue weighs 147 N in vacuum and 139 N when immersed in saltwater of density 1024 kg m-3 . What is the density of the “gold”?
Flotation
For object of uniform density
Fluid DynamicsThe study of fluids in motion.
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.
laminar
turbulent
Equation of continuityStreamlines
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)
tube of flow
Bernoulli’s equation (Daniel Bernoulli, 1700-1782)
For special case of fluid at rest (Hydrostatics!)
For special case of height constant (y1=y2)
The pressure of a fluid decreases with increasing speed
p1 1
2v1
2 gy1 p2 1
2v2
2 gy2
p 1
2v2 gy constant
p2 p1 g(y2 y1)
p1 1
2v1
2 p2 1
2v2
2
Proof of Bernoulli’s equation
Use work energy theoremwork done by external force (pressure)
=change in KE + change in PE
W=K+ UWork done
Change in KE
Change in PEU Vgy2 Vgy1
K 1
2mv2
2 1
2mv1
2 1
2Vv2
2 1
2Vv1
2
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
W p1V p2V (p2 p1)V
ProblemTitanic 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”