Download Physics 201 Chapter 13 Lecture 1

Physics 201
Chapter 13
Lecture 1


Fluid Statics
 Pascal’s Principle
 Archimedes Principle (Buoyancy)
Fluid Dynamics
 Continuity Equation
 Bernoulli Equation
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Fluids
Density = Mass/Volume
Pressure (P)
P = Force/Area [N/m2]
ρ=M/V
units = kg/m3
1 N/m2 = 1 Pascal (Pa)
Pressure variation with depth
P=ρgh
Atmospheric Pressure
Even when there is no breeze air molecules are continuously
bombarding everything around - results in pressure
normal atmospheric pressure = 1.01 x 105 Pa (14.7 lb/in2)
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Compressiblity

Density & Pressure are related
by the Bulk Modulus
 LIQUID: incompressible
(density almost constant)
 GAS: compressible (density
depends a lot on pressure)
Δp
B=
(−ΔV / V )
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Variation of pressure with depth
m = ρV; V = Ah ⇒ m = ρ Ah
ρ Ah ) g
F mg
(
P= =
; i.e., P =
⇒ P = hρ g
A
A
A
True for all shapes of containers
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Pascal’s Principle


A change in pressure in an enclosed fluid is
transmitted undiminished to all the fluid and to its
container.
This principle is used in hydraulic system
 P1 = P2
 (F1 / A1) = (F2 / A2)
F2
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Pascal’s Principle

This principle is used in hydraulic system
 P1 = P2
 (F1 / A1) = (F2 / A2)
 Can be used to achieve a mechanical advantage
 F2 = F1 (A2 / A1)
» Work done is the same: height by which the
surface A2 rises is smaller than the change in
the height of surface with area A1.
F1
A1
A2
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F2
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Using Fluids to Measure Pressure
• Use Barometer to measure Absolute Pressure
 Top of tube evacuated (p=0)
 Bottom of tube submerged into pool of mercury
open to atmosphere (p=p0)
 Pressure dependence on depth:
h=
Barometer
p0
ρg
•
Use Manometer to measure Gauge Pressure
 Measure pressure of volume (p1) relative to the
atmospheric pressure (≡ gauge pressure )
p1
Manometer
Δh
 The height difference (Δh) measures the gauge
( p1 − p0 )
pressure:
Δh =
1 atm = 760 mm (29.9 in) Hg
ρg
= 10.3 m (33.8 ft) H20
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p0
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Measurement of Pressure

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Manometer
 If both sides of an U-tube are open to atmosphere
the levels of the fluid are the same on both sides
 If one side is connected to a “pressurized side” the
level difference between the two sides can be
used to measure pressure.
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Measuring the tire pressure:
Is this a manometer or a barometer?
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Measuring Blood Pressure


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Blood pressure is quite high, 120/80 mm of Hg
Use higher density fluid in a manometer: Mercury
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Archimedes
Object immersed in a fluid is subject to a “buoyant force”.
Force on sides cancel
Force on top Ft = ρghT A
Force on bottom Fb = ρghB A
ΔF = ρg A Δh
FB = (mg)disp
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Archimedes
Object immersed in a fluid is subject to a “buoyant force”.
Force on sides cancel
Force on top Ft = ρghT A
Force on bottom Fb = ρghB A
ΔF = ρg A Δh
FB = (mg)disp
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Float
Weight of object = ρ0gV
Buoyant force is the weight
of the displaced fluid
Weight of fluid = ρfgV
Displace just enough fluid such that forces = 0!
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Archimedes Principle
Buoyant Force (B)
 weight of fluid displaced (P=F/A, P=ρgh)
» B = ρfluid g Vdisplaced
» W = ρobject g Vobject
» object sinks if ρobject > ρfluid
» object floats if ρobject < ρfluid
» Eureka!
If object floats….
» B=W
» Therefore ρfluid g Vdisplaced = ρobject g Vobject
» Therefore Vdisplaced/Vobject = ρobject / ρfluid
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Float
Buoyant force is the weight of the
displaced fluid
Weight of object = ρIceVtotal g
Weight of fluid = ρSeaWatergVsubmersed
Displace just enough fluid such that forces = 0!
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Archimedes Principle
The weight of a glass filled to the brim with water is Wb. A cube of ice
is placed in it, causing some water to spill. After the spilled water is
cleaned up, the weight of the glass with ice cube is Wa. How do the
weights compare:
1. Wb > Wa.
2. Wb < Wa.
3. Wb = Wa.
Archimedes’ Principle: The buoyant force on an
object equals the weight of the fluid it displaces.
Weight of water displaced = Buoyant force =
Weight of ice
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Question
Suppose you float a large ice-cube in a glass of water, and that after
you place the ice in the glass the level of the water is at the very
brim. When the ice melts, the level of the water in the glass will:
1. Go up causing the water to spill.
2. Go down.
3. Stay the same.
Archimedes’ Principle: The buoyant force on an
object equals the weight of the fluid it displaces.
Weight of water displaced = Buoyant force =
Weight of ice
When ice melts it will turn into water of same
volume
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Buoyancy

Two cups hold water at the same level. One of
the two cups has plastic balls (projecting above
the water surface) floating in it. Which cup
weighs more?
Cup I
Cup II
1) Cup I
2) Cup II
3) Both the same
 Archimedes principle tells us that the cups weigh the same.
 Each plastic ball displaces an amount of water that is exactly
equal to its own weight.
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Sunken Balls
Two identical glasses are filled to the same level with water. Solid steel balls are
at the bottom in one of the glasses. Which of the two glasses weighs more?
1. The glass without steel balls
2. The glass with steel balls
3. Both glasses weigh the same
The steel balls sink. The buoyant force equal to the weight of the
displaced water is not sufficient to counter the weight of the steel
balls. Therefore, the glass with steel balls weighs more.
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Buoyant force and depth
Imagine holding two identical bricks under water. Brick A is just beneath the
surface of the water, while brick B is at a greater depth. The force
needed to hold brick B in place is:
1. larger
2. the same as
3. smaller
than the force required to hold brick A in place.
The buoyant force on each brick is equal to the weight of the water it
displaces and does not depend on depth.
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Fluid Flow
Fluid flow without friction
• Volume flow rate: ΔV/Δt = A Δd/Δt = Av (m3/s)
• Continuity: A1 v1 = A2 v2
i.e., flow rate the same everywhere
e.g., flow of river
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Problem
Two hoses, one of 20-mm diameter, the other of 15-mm diameter are
connected one behind the other to a faucet. At the open end of the hose, the
flow of water measures 10 liters per minute. Through which pipe does the
water flow faster?
1. The 20-mm hose
2. The 15-mm hose
3. Water flows at the same speed in both cases
4. The answer depends on which of the two hoses comes first in the flow
When a tube narrows, the same volume occupies a greater length.
For the same volume to pass through points 1 and 2 in a given time,
the velocity must be greater at point 2. The process is reversible.
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Faucet
A stream of water gets narrower as it falls
from a faucet (try it & see).
A1
V1
V2
A2
The velocity of the liquid increases as the water falls due to gravity.
If the volume flow rate is conserved, them the cross-sectional area
must decrease in order to compensate
The density of the water is the same no matter where it is in space
and time, so as it falls down and accelerates because of gravity,the
water is in a sense stretched, so it thins out at the end.
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Streamlines
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Continuity equation
Volume Flow rate
Mass flow rate
In steady state
IV =
IM1 =
Δm1 = ρ1ΔV1 = ρ1 Av1Δt
ΔV
= Av
Δt
Δm1
= ρ1 Av1
Δt
Δm1 Δm2
=
Δt
Δt
General case: mass may be accumulated or decreased in the volume between A1 and A2
IM 2 − IM1 =
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dm2 dm1 dm12
−
=
dt
dt
dt
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Continuity equation
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Bernoulli’s Equation
Pressure drops in a rapidly moving fluid
 whether or not the fluid is confined to a tube
For incompressible, frictionless fluid:


1 2
P + ρv + ρ gh = constant
2
1 2 1 2 1 KE
ρv = mv
=
2
2
V
V
mgh PE
ρ gh =
=
V
V
Bernoulli equation states conservation of energy
For Static Fluids:P1 + ρ gh1 = P2 + ρ gh2
1 2
1 2
Bernoulli's Principle (constant depth):P1 + ρv1 = P2 + ρv2
2
2
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