Transcript Fluid Mechanics - Damien AP Physics

Fluid Mechanics
Fluids flow!
What do you remember about
liquids and gases from chem?
Liquids and Gases
• Liquids:
• Molecules close
together so interact
• Fixed volume
• Variable shape
• Gases
• Molecules far apart
so almost no
interaction
• Variable volume
• Variable shape
Density: mass/volume
• In fluids, density more
useful than mass
• Denoted by Greek
letter rho:
• =m/V
• Also seen as how it
relates to the density of
water: specific gravity
• Sp gr= substance/ water
• If > 1 then denser than
water
Pressure: Force/Area
• Pressure is force
perpendicular to the
surface/area of contact
• Unit = Pascal (Pa) =
N/m2
• 101300Pa= 1atm
Hydrostatic Pressure
• But, since we like to use density instead of mass
for a fluid, replace mass with V so force= Vg
• Pressure= Vg/A
• But just take the V/A: lwh/lw so V/A=h
• THUS: P=gh Shape of container doesn’t matter
Pressure Increases with depth
• P=P0+gh
• P0 for open container
=Patm
• Often problems will ask
for gauge pressure
which is P-Patm
• Pressure directly related
to depth
• Why are the metal bands
closer together at the
bottom of the water
tower?
Direction of Pressure
• Pressure from a fluid acts
perpendicular to the
surface- whichever way
that surface is pointing!
• A confined fluid transmits
externally applied pressure
uniformly in all directions
• REMEMBER- magnitude
depends only on depth
What does this mean in
practice?
• Scuba divers will feel pressure on all sides
Pressure Problem 1
• What is the hydrostatic pressure at a
point 10m below the surface of the
ocean? Sea water has a specific
gravity of 1.025.
Solution 1
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•
Sp gr= substance/ water
seawater=1025kg/m3
Pgauge=gh=(1025kg/m3)(10N/kg)(10m)
=1.025x105Pa
Pressure Problem 2
• Remember when you
were little and you used
to put your finger on the
top of a straw to lift
some of your drink out?
• Consider the air
pressure inside the
straw- above the drink
but below your finger- is
it same, greater, or less
than the atmospheric
pressure?
• Discuss with your
neighbor
Solution 2
P=1atm
Or: PV=nRT you may have
noticed that a little bit of fluid
spills out when you do thisincreased volume means less
pressure
• The pressure is…LESS
THAN atmospheric!
• Why?
• The pressure at the
bottom of the fluid is
equal to atmospheric (it
is a surface that is
exposed to the
atmosphere
• So P=P0+ gh….solving
for P0=P- gh so the
pressure at the top is
the atmospheric
pressure minus
something!
Problem 3
• A flat piece of wood of area 0.5m2 is
lying on the bottom of a lake. The depth
of the lake is 30m. What is the force on
the wood due to pressure?
• Use Patm=1x105Pa
• Use water=1000kg/m3
Solution 3
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•
•
•
•
P= P0+gh
P=1x105Pa+ (1000kg/m3)(10N/kg)(30m)
P=4x105Pa
Pressure=Force/Area so F=PA
F=(4x105Pa)(0.5m2)
Buoyancy
Buoyancy
• How can you tell
whether an object
will float or sink?
• For the group of
objects on the desk,
what will determine
whether they float or
sink?
Buoyancy
• Pressure at the
bottom of an
object
submerged in a
fluid is greater
because it is at a
greater depth
• Buoyant force is
due to pressure
differences top to
bottom- does
not depend on
depth
• The strength of the
buoyant force is equal
to the weight of the
fluid displaced by the
object
• When object is partially
or completely
submerged in fluid, the
volume submerged=the
volume of water
displaced=Vsub
• Weight =mg so:
• Fbuoy=fluidVsubg
Archimedes’
Principle
Floating, Sinking
• When an object floats, the buoyant force it
feels balances its weight
• Vsub/V=object/fluid
• So if object<fluid object will float
• The fraction of its volume that will submerge
= the ratio of its density to the fluid’s density
Submerged
• If the object is the
same density as the
fluid, it will happily
hover anywhere
under the surface
Problem-Solving: Buoyancy
• An object with mass of 150kg and a
volume of 0.75m3 is floating in ethyl
alcohol whose specific gravity is 0.8.
What fraction of the object’s volume is
above the surface of the fluid?
Solution: buoyancy
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•
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•
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object=m/V=150kg/0.75m3=200kg/m3
The density of the fluid=(sp gr)(water)
fluid=(0.8)(1000kg/m3)=800kg/m3
object/fluid=200/800=0.25
So 25% or 1/4 of the object is below
surface so 3/4 or 75% is above
Flow Rate
• Flow rate, f, is the
volume that passes a
point per unit of time
• SI unit: m3/s
• f=cross section
area*speed of flow
• f=Av
• Careful- don’t confuse
flow rate with flow
speed: flow rate is how
much fluid per time
Continuity Equation
• If a pipe is carrying a
liquid, the flow rate
must be the same
throughout the pipe:
f1=f2 so:
• A1v1=A2v2
• So fluid speed
inversely proportional
to area
• Thus the thumb-overnozzle effect on a
garden hose
Continuity Problem 1
• A pipe carrying water has a radius of
2cm and a flow speed of 6m/s
• What is the flow rate?
• What is the flow speed if the pipe
constricts to a 1cm radius?
Continuity Solution 1
• f=Av
• f=(r2)v=(0.02m)2(6m/s)=0.0075m3/s
• By continuity equation, we know f must stay
constant so if radius decreases by a factor of
2, the area must decrease by a factor of 4
(A=r2). So the flow speed must increase by
a factor of 4. 6m/s*4=24m/s
Continuity Problem 2
• If the diameter of a pipe increases from
4 to 12 cm, what will happen to the flow
speed?
Continuity Solution 2
• If diameter increases by a factor of 3 (so
will radius), the area increases by a
factor of 32 so the velocity will decrease
by a factor of 32=9.
Ideal Fluid Flow
• Fluid is incompressible
– Works for liquids and if pressure constant
works for gases too
• Viscosity is negligible
– So works for water but not maple syrup
• Flow is streamline- moving smoothly
through tube (as opposed to turbulent)
Bernoulli’s Principle
• Conservation of energy
applies to moving fluids:
• Pressure + potential
energy +kinetic energy is
constant
• p+gy+1/2v2 is
constant
• So as v increases,
pressure drops!
So what?
• It’s how airplanes
fly!
• Air moves faster
over curved top of
wing- increased v
means decreased
pressure creating lift
So what?
• It’s how we get baseballs to curve, lift, drop