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Transcript of Ch15 testing
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Chapter 15: Fluid Motion
Fluids Characteristics of fluids
• Microscopically molecules of a fluid do not have long-range
order. But liquids do have short-range order unlike gases
• Fluids can flow and conform to the boundaries of a container
• Fluids cannot sustain a shearing stress
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Fluids (cont’d)
Some useful quantities
Density for
V
m
V
m
for uniform solid or liquid
unit kg/m3 (at STP for a gas)
333
kg/m2.1;kg/m100.1 air water
3333 kg/m1035.11;kg/m101.0 lead styrofoam
Pressure for area
m V mass ,volume
A and normal force F
dA
dF
A
F p
A
0
unit pascal2m1/ N1Pa in SI unit
in SI unit
other useful units:
25 in/lb7.14torr 760Pa1001.1atm1
height of a column of Hg corresponding to this pressure (mm)
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Pressure
Pressure of a liquid at rest (uniform density)
A
A
2 y
1 y
fluid level
imaginary
box
2 F
1 F
mg
A p F 11
A p F 22
0)( 1221 gA y y F F
weight of the imaginary box
hg
g y y p p
)( 1212
mg
h
hg p p 12
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Pressure (cont’d)
Pressure of a fluid at rest : gauge pressure
A
A
2 y
1 y
fluid level
imagi
nary
box
2 F
1 F
mg
h
hg p p 12
0 p
gauge pressure:
gauge
gauge
p p p
p p p
202
101
atmospheric pressure
atmospheric
pressure
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A simple model for atmospheric pressure
Pressure of a gas at rest ( )kp
gdy y y g p pdp )( 2112
When ,0, 1212 y y p p
kpgdy gdydp
kpg dydp /
)(
1212 y ykg e p p
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Barometer Pressure measurement using a liquid:
(measures absolute pressure)
h
0 p
0 p
atmospheric
pressure
From )( 2112 y y g p p
as 0122 ,,0 p ph y p
gh p 0
If the liquid is mercury, for 1 atm :
m/s83.9,kg/m106.13
,Pa103.101atm1
33
3
0
g
p
Hg
mm760m758.0/0 g ph
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Manometer
Pressure measurement using a liquid :
(measures gauge pressure)
g p h
liquidtank
manometer
0 p
level 1
level 2
)( 2112 y y g p p
0,
,,
101
22
y p p
h y p p
gh p p p g 0
gauge pressure
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Pascal’s law
Pressure applied to an enclosed fluid is transmitted
undiminished to every portion of the fluid and walls
of the containing vessel
piston
weight
liquid
P
h
ext p
p
(area A)
w
gh p p pext
0
pressure at P
pressure due
to weight w:
w/A
atmospheric
pressure
pressure due to
liquid above P
incompressible
ext p p
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Pascal’s law (cont’d)
Hydraulic lever
1 x
2 x1 F
2 F
1 A2 A
2211// A F A F p
1122 )/( F A A F
1212 1/ F F A A
12122211 )/( x A A x A x A x
111
121112222 )/()/(
W x F
x A A F A A x F W
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Buoyancy
Origin of buoyancy
• Consider a submerged massless object filled with the same fluid
as the fluid that surrounds the object
1 P
2 P
1 F
2 F
0)( 12 g m F F F f net
buoyancymass of fluid in the object
The object is at rest
• Now fill the object with another material
g m F f b
buoyant force
0)( 12 g m F F F objnet
g m F f b
opposite dir.
mf g
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Buoyancy
Origin of buoyancy (cont’d)
• Consider a portion of fluid at rest in a container surrounded by animaginary boundary represented in dashed line.
• Since the portion of the fluid defined by the surface in dashed line
is at rest, the net force on this portion due to pressure must be
equal to that of the weight of the fluid inside the surface, and
opposite in direction.
dF
dF
dF
dF
dF
dF
dF
dF
• The same argument can be applied when the
imaginary portion of the fluid is replaced by an
object that occupies the same space.
X
0fluid bfluidnet g V F g m F d F
Fnet mg
Vg g m F fluidfluid b buoyant force
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Buoyancy
Archimedes’s principle
• The buoyant force on a partly or completely submerged objectis equal to the weight of the displaced fluid:
Vg
g m g m g m F F
obj f
obj f objbnet
)(
• If
0, net obj f F
apparent weight g mVg W objobj f
)(
0, net obj f F
• If
The object will rise until a part of it comes out above the fluid
Surface when the average density increases to f
Object sinks
Object floats
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What fraction of an iceberg is submerged in the sea water?
Let’s assume that the total volume of the iceberg is Vi.
Then the weight of the iceberg Fgi is
Since the whole system is at its
static equilibrium, we obtain
g V F ii gi
Let’s then assume that the volume of the iceberg
submerged in the sea water is Vw. The buoyant force B
caused by the displaced water becomes
g V B ww
g V g V wwii
Therefore the fraction of the
volume of the iceberg
submerged under the surface of the sea water is
890.0/1030
/9173
3
mkg
mkg
V
V
w
i
i
w
About 90% of the entire iceberg is submerged in the water!!!
Example
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• A fake or pure gold crown?
Is the crown made of puregold?
Tair =7.84 N
Twater =6.86 N0mg T air
0 Bmg T water
0 BT T air water
N980.0
water water water air gV T T B
34 m1000.1 crownwater V V
kg800.0/ g T m air
33 kg/m1000.8/ crowncrown V m
gold=19.3x103 kg/m3
Example
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• Floating down the river
What depth h is the bottom of the raft submerged?
A=5.70 m2 wood=6.00x102 kg/m3
g m B g m B raft raft 0
g Ah g V g m B water water water water )()(
g V g m raft raft raft )(
g V g Ah raft raft water )()(
m0632.0 A
V h
water
raft raft
Example
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Density and Pressure
Example
• Oil and water
=0.700 g/cm3
=1025 kg/m3
h1=8.00 m
h2=5.00 mPa101.56
m)00.8(
)m/s80.9)(kg/m1000.7(
Pa1001.1
5
232
5
101
gh P P
Pa1006.2 5
21
gh P P bot
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Ideal fluid flow
Ideal fluids in motion
• Incompressible ( density is constant at any position)• No internal friction (no viscosity)
• Steady (non-turbulent) flow- the velocity at a point is
constant in time.
flow tube
flow lines
flow line : The path of an individual particle in a
moving fluid
steady flow: A flow whose pattern does not change
with time. Every element passing
through a given point follows the same
flow line
streamline : A curve whose tangent at any point isin the direction of the fluid velocity at
that point
flow tube : The flow lines passing through the edge
of an imaginary area such as A A
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Continuity equation
Continuity equation I (incompressible fluid)
1 A
2 A
dt v1
dt v2
1v
2v
Avdt dV /
The volume of the fluid that passes through
area A during a small time interval dt :
Avdt dV
dt v AdV dt v AdV 222111 ;
•In an ideal fluid the density is constant.
•In a time interval dt the mass that flowsinto Area 1 is the same as the mass that
flows out of Area 2.
22112211
v Av Adt v Adt v A
The mass of a moving fluid does not change as it flows.
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Continuity equation
Volume flow rate
1 A
2 A
dt v1
dt v2
1v
2v
Avdt dV
222111222111 v Av Adt v Adt v A
Avdt dV / volume flow rate
Continuity equation II
(compressible fluid)
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Bernoulli’s equation
Work done by pressure
1 A
2 A
dt vds 11
dt vds 22
1v
2v222 A p F
111 A p F
1 y2 y
dV
dV
dV p p
ds A pds A pdW
)( 21
222111
Change in kinetic energy
2211 ds Ads AdV
)()2/1( 2
1
2
2 vvdV dK
)( 12 y ydVg dU
Change in potential energy
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Bernoulli’s equation
dU dK dW
Energy conservation
dV p pdW )( 21
)()2/1( 2
1
2
2vvdV dK
)( 12 y ydVg dU
dV p p )( 21 )()()2/1( 12
2
1
2
2 y ydVg vvdV
)()()2/1( 12
2
1
2
221 y y g vv p p
const.)2/1()2/1(2
222
2
111 v gy pv gy p
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Example
Venturi meter
2
22
2
112
1
2
1v pv p
)1(2
12
2
2
12
121 A
Av p p
1
2
12 v
A
Av
h
gh p p 21
1)/(
22
21
1
A A
ghv
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Example Torricelli’s theorem
The velocity of the fluid coming out of a hole in a tank as
shown in the figure can be calculated using Bernoulli’s
equation.
At the top surface the velocity
of the fluid is zero. The pressures
at the top surface and at the hole
are the same, namely, the
atmospheric pressure.
22 )2/1()2/1( holeholeholetoptoptop v gy pv gy p
ghvv y y g holeholeholetop
2)2/1()( 2
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Example Siphon
Suppose a U-shaped piece of pipe is completely submerged in
water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of
water, the water does not run out of the pipe. WHY?
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Example Siphon
Suppose a U-shaped piece of pipe is completely submerged in
water, filled with water, and then turned upside down under water. As you slowly pull the top of the U-shaped piece of pipe out of
water, the water does not run out of the pipe. WHY?
Air cannot enter the pipe. As the water starts running out of the pipe,
a near vacuum is created in the topmost region of the inverted U. The
pressure here drops to near zero. The atmospheric pressure on the
surface of the water in the bucket pushes the water into the U-shaped
pipe.
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Example Siphon
If a U-shaped hose or pipe connects a liquid-filled container at a
higher altitude to a container at a lower altitude over a barrier, theliquid can be siphoned into the container at the lower altitude.
Atmospheric pressure helps to push the liquid over the barrier.
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Example Siphon
When P1>P2, the fluid can be siphoned from the left to the
right bucket.
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• Water garden
A water hose 2.50 cm in diameter is used by a gardener to fill a
30.0-liter bucket. The gardener notices that it takes 1.00 min to fillthe bucket. A nozzle with an opening of cross-sectional area 0.500
cm2 is then attached to the hose. The nozzle is held so that water
is projected horizontally from a point 1.00 m above the ground. Over
what horizontal distance can the water be projected?
/sm1000.5s0.60
min00.1
cm100.0
m00.1
L00.1
cm1000.1
min1.00
L0.30rateflowvolume 33
333
velocity)initialtheof component-xthe:( 0022211 x x vv Av Av A
m/s0.10m10500.0
/m1000.524
34
2
110
s
A
v Av x
m52.4s;452.0
m/s80.9
m)00.1(22
2
102
2
0
t v x
g
t t gt t v y x y
Example
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Consider a water tank with a hole.
(a)Find the speed of the water
leaving through the hole.
• Example : A water tank h =0.500 m
y1 =3.00 m
201
2
10
2
1 gy P gyv P
m/s13.3
2)(2 121
gh y y g v
(b) Find where the stream hits the ground.
s782.02
10 0
2
1 t t v gt y y y
m45.210 t vt v x x
x
y
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Find the speed at Point 1.
• Example : Fluid flow in a pipe
A2=1.00 m2 A1=0.500 m2
h =5.00 m
2
2
201
2
102
1
2
1 gyv P gyv P
2211 v Av A
1
2
12 v
A
Av
2
2
1
2
101
2
102
1
2
1 gyv A
A P gyv P
m/s4.11
)/(1
22)(21
2
21
112
2
2
12
1
A A
ghv gh y y g
A
Av
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Viscosity and turbulence
Viscosity
Viscosity is internal friction in a fluid, and viscous forces opposethe motion of one portion of a fluid relative to another.
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Viscosity and turbulence
Drag
If a fluid in laminar flow flows around an obstacle, it exerts a viscousdrag on obstacle. Frictional forces accelerate the fluid backward
against the direction of flow and the obstacle forward in the direction
of flow.
laminar flow
adjacent layers of fluid slide
smoothly past each other andflow is steady
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Viscosity and turbulence
Turbulence
When the speed of a flowing
fluid exceeds a certain critical
value the flow is no longer laminar.
The flow patter becomes extremely
irregular and complex, and it changes
continuously in time. There is nosteady flow pattern. This chaotic flow
Is called turbulence.
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Problems
Problem 1
2.00 mThe upper edge of a gate in a dam runsthe water surface. The gate is 2.00 m high
and 4.00 m wide and is hinged along the
horizontal line through its center. Calculate
the torque about the hinge arising from
the force due to the water.
Solution
Denote the width and depth at the bottom of the gate by w and H.
The force on a strip of vertical thickness dh at a depth h is:
and the torque about the hinge is After integrating from h=0 to h=H, you get the torque:
)(wdh ghdF
.)2/( dh H h gwhd
.1061.212/43
m N gwH
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Problem 2
Solution
(a) From Archimedes’s principle so
An object with height h, mass M, and a uniform cross-sectional area A
floats upright in a liquid with density .
(a)Calculate the vertical distance from the surface of the liquid to thebottom of the floating object in equilibrium.
(b) A downward force with magnitude F is applied to the top of the object.
At the new equilibrium position, how much farther below the surface of
the liquid is the bottom of the object than it was in part (a)?
(c) Calculate the period of the oscillation when the force F is suddenlyremoved.
, Mg gLA )./( A M L
(b) The buoyant force is: With the result of part (a) solvingfor x gives:.)( F Mg x L gA
)./( gA F x
(c) The force is always in the direction toward the equilibrium, namely, a
restoring force Therefore the “spring constant” is
and the period of the oscillation is
. gAx Mg F F buoyres gAk .)/(2/2 gA M k M T
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Problem 3
Solution
You cast some metal of density m in a mold, but you are worried that there
might be cavities within the casting. You measure the weight of the casting
to be w, and the buoyant force when it is completely surrounded by water
to be B. (a) Show that is the total volume
of any enclosed cavities. (b) If your metal is copper, the casting’s weight
is 156 N, and the buoyant force is 20 N, what is the total volume of any
enclosed cavities in your casting? What fraction is this of the total volume
of the casting?
(a) Denote the total volume V. If the density of air is neglected, the buoyant
force in terms of the weight is:
Therefore
]/)/[( 0V g w g gV B mwater water
)/()/(0 g w g BV mwater
)./()/(0 g w g BV mwater
(b) .1052.2)/()/( 34 m g w g B Cuwater
The total volume of the casting is
),/( g B water The cavities are 12.4% of the total volume.
P bl 4
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Problem 4
Solutions
A U-shaped tube with a horizontal portion of
length contains a liquid. What is the difference
in height between the liquid columns in the
vertical arms (a) if the tube has an accelerationtoward the right? (b) if the tube is mounted on
a horizontal turntable rotating with an angular
speed with one of the vertical arms on the
axis of rotation?
a
(a)Consider the fluid in the horizontal part of the tube. This fluid with mass
, is subject to a net force due to the pressure difference between
the ends of the tube, which is the difference between the gauge pressures
at the bottoms of the ends of the tubes. Now this difference is
and the net force on the horizontal part of the fluid is
or
(b) Similarly to (a) consider the fluid in the horizontal part of the tube. As in (a)
the fluid is accelerating. The center of mass has a radial acceleration of
magnitude so the difference in heights between the columns
is
),( R L y y g
A
,)( a A A y y g R L .)/()( g a y y R L
,2/2 rad a
).2/()/)(2/( 222 g g
A A