Chapter 14. Fluid Mechanics
Dr. Yousef Abou-Ali
yabouali@iust.edu.sy
Syllabus
14.1 Density
14.2 Pressure in a Fluid
14.3 Buoyancy
14.4 Fluid Flow
14.5 Bernoulli's Equation
14.6 Viscosity and Turbulence
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Goals for Chapter 14
• To study density and pressure in a fluid.
• To apply Archimedes Principle of buoyancy.
• To describe surface tension and capillary action.
• To study and solve Bernoulli's Equation for fluid
flow.
• To see how real fluids differ from ideal fluids
(turbulence and viscosity).
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University Physics, Chapter
14
Introduction
• Fluids play a vital role in many aspects of everyday life.
• We drink them, breathe them, and swim in them.
• They circulate through our bodies and control our weather.
• Airplanes fly through them; ships float in them.
• A fluid is any substance that can flow; we use the term for both
liquids and gases.
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University Physics, Chapter
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Introduction
• We
usually think of a gas as easily compressed and a liquid as
nearly incompressible.
• We begin our study with fluid statics, the study of fluids at rest in
equilibrium situations.
• Than will move to talk about fluid dynamics.
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University Physics, Chapter
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14.1 Density
• An important property of any material is its density, defined as its
mass per unit volume.
• A homogeneous material such as ice or iron has the same density
throughout.
• We use the Greek letter ρ (“rho”) for density.
• If a mass m of homogeneous material has volume V , the density ρ
is:
m
ρ =
V
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(d efin ition of d en sity )
(14.1)
University Physics, Chapter
14
14.1 Density
• At
right, liquids of different
densities separate with denser
liquids lower in the glass.
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University Physics, Chapter
14
14.1 Density
• The density of some materials varies from point to point within the
material; some examples are the earth’s atmosphere,
which is less dense at high altitude.
and oceans (which are denser at greater depths).
• For these materials, Eq. (14.1) describes the average density.
• In general, the density of a material depends on environmental
factors such as temperature and pressure.
• The SI unit of density is the kilogram per cubic meter (1 kg/m3).
3
3
1 g/cm = 1000 kg/m
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University Physics, Chapter
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14.1 Density
• The specific gravity of a material is the ratio of its density to the
density of water at 4.0o C, 1000 kg/m3.
• It is a pure number without units.
• Since it has nothing to do with gravity; “relative density” would
have been better.
Example 14.1 (The weight of a roomful of air):
Find the mass and weight of the air in a living room with a
4.0
m × 5.0 m floor and a ceiling 3.0 m high. What is the mass and
weight of an equal volume of water?
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.1 Density
Solution:
• Identify : We assume that air is homogeneous, so that the density is
the same throughout the room.
• Set Up : We use Eq. (14.1) to relate the mass (the target variable) to
the volume (which we calculate from the dimensions of the room)
and the density (from Table 14.1, in your book).
• Execute: The volume of the room is:
3
V = (3.0m)(4.0m)(5.0m) = 60 m
The mass m air of air is given by Eq. (14.1):
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.1 Density
3
3
mair = ρair V = (1.2 kg / m )(60m ) = 72 kg
• The weight of the air is:
2
wair = mair g = (72 kg) (9.8 m/s ) = 700 N
• The mass of an equal volume of water is:
mwater = ρ water V = (1000 kg / m 3 ) (60 m 3 ) = 6.0 × 10 4 kg
• The weight is:
wwater = mwater g = (6.0 × 10 4 kg) (9.8 m/s 2 ) = 5.9 × 105 N
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
• We define the pressure p at the point as the normal force per unit
area, that is, the ratio dF┴ to dA :
d F⊥
p =
dA
(d e f in itio n o f p r e s s u r e )
(1 4 .2 )
• If the pressure is the same at all points of a finite plane surface
with area A , then:
F⊥
p=
A
(14.3)
• The SI unit of pressure in the pascal, where:
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1 pascal = 1 Pa = N/m2
University Physics, Chapter
14
14.2 Pressure in a Fluid
• Two
related units, used principally in meteorology, are the bar ,
equal to 105 Pa, and millibar , equal to 100 Pa.
• Atmospheric pressure p a is the pressure of the earth’s atmospheric,
the pressure at the bottom of this sea of air which we live.
( pa )av = 1atm = 1.013 ×105 Pa
= 1.013 bar = 1013 millibar
Example 14.2 ( The force of air):
In the room described in Example 14.1, what is the total downward
force on the surface of the floor due to air pressure of 1.00 atm?
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Solution:
• Identify and Set Up : The pressure is uniform, so we use Eq. (14.3)
to determine the force from the pressure and area.
• Execute: The floor area is A
= (4.0 m) (5.0 m) = 20 m2, so from
Eq. (14.3) the total downward force is
5
2
2
F⊥ = p A= (1.013 × 10 N/m )(20 m )
= 2.0 × 106 N
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• If the weight of the fluid can be neglected, the pressure in a fluid is
the same throughout its volume. But often the fluid’s weight is not
negligible.
• Atmospheric
pressure is less at high altitude than at sea level,
which is why an airplane cabin has to pressurized when flying at
35.000 feet.
• When you dive in deep water, your ears tell you that the pressure
increases rapidly with increasing depth below the surface.
• We can derive a general relation between the pressure p
at any
point in a fluid at rest and the elevation y of the point.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
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14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• We
will assume that the density ρ and the acceleration due to
gravity g are the same throughout the fluid.
• If the fluid in equilibrium, every volume element is in equilibrium.
• Consider a thin element of fluid with height dy (figure below). The
bottom and top surface have area A , they are at elevations y &
y + dy , above some reference level where y = 0.
• The volume of the fluid element is dV = A dy , its mass is dm = ρ dV
= ρ A dy , and its weight is dw = dm g = ρgA dy .
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University Physics, Chapter
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14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
(p + dp) A
dy
A
dw
pA
y
Element of fluid
thickness dy
(b)
0
(a)
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• What are the other forces on this fluid element?
The pressure at the bottom surface p ; the total y -component of
upward force on this surface is pA .
The pressure at the top surface is p + dp , and the total
y -component of (downward) force on the top surface is – (p + dp )A .
• The fluid element is in equilibrium, so
∑ Fy = 0 so pA − ( p + dp ) A − ρ gA dy = 0
dp
= −ρ g
dy
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(14.4)
University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• If p 1 and p 2 are the pressure at elevations y 1 and y 2, respectively,
and if ρ and g are constant, then
p2 − p1 = −ρ g ( y2 − y1)
( pressure in a fluid of uniform density) (14.5)
• It is often convenient to express Eq. (14.5) in terms of the depth
below the surface of a fluid (figure below).
Fluid, density ρ
2
P2=
y2 – y1 =
P1 =
Dr. Y. Abou-Ali, IUST
y2 – y1
y1
1
y2
University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
po − p = − ρ g ( y2 − y1) = − ρ gh
p = po + ρ g h
( pressure in a fluid of uniform density ) (14.6)
• The pressure p at a depth h is greater than the pressure p o at the
surface by an amount ρgh.
• Note that the pressure is the same at any two points at the same
level in the fluid. The shape of the container does not matter.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• The pressure in any fluid at the same elevation will be the same
regardless of the shape or size of the container.
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University Physics, Chapter
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14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
Pascal’s law: Pressure applied to an enclosed fluid is transmitted to
every portion of the fluid and the walls of the containing vessel.
• The hydraulic lift shown schematically in figure below illustrates
Pascal’s law.
• A piston with small cross-sectional area A 1 exerts a force F1 on the
surface of a liquid such as oil. The applied pressure p = F1/A 1
transmitted through the connecting pipe to a larger piston of area
A 2.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Pressure, Depth, and Pascal’s Law
• The applied pressure is the same in both cylinders, so:
F1 F2
p=
=
A1 A2
&
A2
F2 =
F1
A1
(14.7)
14.7)
• In a room with a ceiling height 3.0 m filled with air of uniform
density 1.2 kg/m3, the difference in pressure between floor and
ceiling given by Eq. (14.6):
ρgh = (1.2 kg/m3) (9.8 m/s2) (3.0 m) = 35 Pa
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University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• If the pressure inside a car tire is equal to atmospheric pressure,
the tire is flat.
• The pressure has to be greater
than atmospheric to support the
car, so significant quantity is the difference between the inside and
outside pressure.
• When we say that the pressure in a car tire is (220 kPa or 2.2 × 10 Pa),
5
we mean it is greater than atmospheric pressure (1.01 × 105 Pa) by
this amount.
• The total pressure in the tire is then 320 kPa.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The excess pressure above atmospheric pressure is usually called
gauge pressure.
• The total pressure is called absolute pressure.
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University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
Example 14.3 ( Finding absolute and gauge pressures):
A solar water-heating system uses solar panels on the roof, 12.0 m
above the storage tank. The water pressure at the level of the panels
is one atmosphere. What is the absolute pressure in the tank? The
gauge pressure?
Solution:
• Identify : Water is nearly incompressible. Hence we can treat it as a
fluid of uniform density.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• Set Up : The level of the panels corresponds to point 2 in the figure
above, and the level of the tank corresponds to point 1. Hence our
target variable is p in Eq. (14.6); we are given p o= 1 atm = 1.01 × 10-5
Pa and h = 12.0 m.
• Execute: From Eq. (14.6), the absolute pressure is:
p = po + ρ gh
= (1.01× 105 Pa )+(1000 kg/m3 )(9.80 m/s 2 )(12.0 m)
= 2.19 × 105 Pa = 2.16 atm
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
The gauge pressure is:
5
p − po = (2.19 − 1.01) ×10 Pa
5
= 1.18 ×10 Pa = 1.16 atm
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The simplest pressure gauge is the open-tube manometer
(figure
below). The U-shaped tube contains a liquid of density ρ.
Pa=
y2 - y1=
y2
Pressure p
y1
P + ρgy1
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Pa + ρgy2
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The pressure at the bottom of the tube due to the fluid in the left
column is p + ρgy 1, and the pressure at the bottom due to the fluid in
the right column is p a + ρgy 2.
• These pressure are measured at the same point, so they must be
equal:
p + ρ gy1 = p a + ρ gy 2
p − pa = ρ g ( y2 − y1 ) = ρ gh
(14.8 )
• In Eq. (14.8), p is the absolute pressure, and the difference p – p a
between absolute and atmospheric pressure is the gauge pressure.
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University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• Another
common pressure gauge is the mercury barometer. It
consists of a long glass tube, closed at one end, that has to be filled
with mercury and then inverted in a dish of mercury.
p0 = 0
p = pa
y2 – y1=
y2
y1
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The pressure p o at the top of the mercury column is practically
zero. From Eq. (14.6),
pa = p = 0 + ρ g ( y2 − y1 ) = ρ gh
(14.9 )
• Thus
the mercury barometer reads the atmospheric pressure p a
directly from the height of the mercury column.
• In
many applications, pressures are still commonly described in
terms of the height of the corresponding mercury column (mm Hg).
• A pressure of 1 mm Hg is called 1 torr.
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• One
common type of blood-pressure gauge,
sphygmomanometer, uses a mercury-filled manometer.
called
a
• Blood-pressure readings, such as 130/80, give the maximum and
minimum gauge pressures in the arteries, measured in mm Hg or
torr.
• Blood pressure varies with height; the standard reference point is
the upper arm, level with the heart.
• We
distinguish high-and low pressure sections of the
cardiovascular system because the blood pressure varies
significantly as illustrated in figure below.
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University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The
high-pressure part includes the aorta, the arteries and
arterioles and the capillaries of the systemic circulation.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• The blood pressure in the arteries varies and 16.0 KPa (systolic
pressure equal to 120 mm Hg).
• The
low-pressure part includes the veins and the pulmonary
circulation in this circulation the pressure varies only between 1.3
KPa and 3.3 KPa (10 - 25 mm Hg).
• The pressure in our cardiovascular system exceeds the ambient air
pressure everywhere, this is correct only while lying down negative
gauge pressures can occur while standing.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
Example 14.4 ( A tale of two fluids):
A manometer tube is partially filled with water. Oil (which does not
mix with water, and has a lower density than water) is then poured
into the left arm of the tube until the oil-water interface is at the
midpoint of the tube (figure below). Both arms of the tube are open
to the air. Find a relationship between the heights hoil and hwater.
Solution:
• Identify : the relationship between pressure and
depth in a fluid
applies only to fluids of uniform density. Hence we cant write a
single equation for the oil and the water together. We can write a
pressure-depth relation for each fluid separately.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
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14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
hoil
hwater
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
• Set Up : Let p o be atmospheric pressure and let p be the pressure at
the bottom of the tube. The densities of the two fluids are ρwater and
ρoil (which is less than water). We use Eq. (14.6) for each fluid.
• Execute: For the two fluids, Eq. (14.6) becomes,
p = po + ρ water g hwater
p = p o + ρ oil g hoil
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Absolute Pressure, Gauge Pressure, and Pressure Gauges
Since the pressure p at the bottom of the tube is the same for both
fluids,
ρ water
hoil =
hwater
ρoil
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University Physics, Chapter
14
14.2 Pressure in a Fluid
Example:
Relative to the blood pressure in
supine position calculate the additional
blood pressure difference between the
brain and the feet in a standing
standard man. Use ρ = 1.06 g/cm3.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.2 Pressure in a Fluid
Solution:
• To
quantify the additional difference for the standing standard
man we use pascal’s law:
• For the pressure difference:
h = ybrain − yfeet
• For the height of the person. this choice
for ∆p eliminates the extra
minus sign on the right hand:
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University Physics, Chapter
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14.2 Pressure in a Fluid
∆p = ρblood gh
• This difference is about 20% of the atmospheric pressure.
• For physiological applications it is more useful to refer to pressure
that are measured relative to the pressure at the height of the heart.
• For a standing person of 1.73 m height the heart is at a height of
1.22 m and the arterial venous pressure in the feet are increased
relative to the pressure at the height of the heart by:
∆p = ρblood gh = (1.06 × 103 )(9.80 )(1.22 ) = 12.7 KPa = 95 mmHg
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University Physics, Chapter
14
14.2 Pressure in a Fluid
• The
pressure is increased to an average arterial value of 190
mm Hg and an average venous value of 100 mm Hg.
• In the scalp the pressures decrease for a standing person by:
∆p = ρblood gh = (1.06 × 103 )(9.80)( −0.51) = −5.3 KPa = -40 mmHg
• The average arterial pressure drops to a value of 55 mm Hg and
the average venous pressure becomes -35 mm Hg.
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Blood Pressure Measurement
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
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Blood Pressure Measurement
• The
figure above, illustrated what the health practitioner hears
with a stethoscope as a function of pressure.
• Each
sketch shows the pressure relative to systolic (ps) and
diastolic (pd) pressure at the right the normal variation of the blood
pressure (red curve) and the audible sound (orange curve).
• Initially (Fig. a) the pressure in the cuff is increased until the flow
of the blood through the brachial artery is below the cuff. A valve on
the bulb is then opened to lower the pressure in the cuff.
• When
the pressure in the brachial artery falls just below the
maximum pressure generated by the heart (which is the systolic
pressure, where systole refers to the contraction of the heart muscle ,
the artery opens momentarily on each beat of the heart .
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University Physics, Chapter
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Blood Pressure Measurement
• The velocity of the blood in the artery is high and the blood flow is
turbulent during these events this leads to a noisy blood flow easily
recognized by the health practitioner. The manometer now reads 120
mm Hg for a standard man with a healthy heart.
• When
the pressure in the cuff is lowered further (Fig.c)
intermittent sounds are heard until the pressure in the cuff falls
below the minimum heart pressure (which is the diastolic pressure
where diastolic refers to the expansion of the heart muscle) the a
continuous background sound is heard as illustrated in (Fig.d) the
transition to the continuous sound occurs at about 80 mm Hg for the
standard man with a healthy heart.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.3 Buoyancy
• Buoyancy is a familiar phenomenon: a body immersed in water
seems to weight less than when it is in air.
• When the body is less dense than the fluid, it floats.
• The human body usually floats in water.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.3 Buoyancy
Archimedes’s principle states: When a body is completely or
partially immersed in a fluid, the fluid exerts an upward force on the
body equal to the weight of the fluid displaced by the body.
• The
upward force exerted on
the body by the fluid is called
buoyant force.
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University Physics, Chapter
14
14.3 Buoyancy
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.3 Buoyancy
Example 14.5 ( Buoyancy):
A 15.0-kg solid gold statue is being raised from a sunken ship (figure
below). What is the tension in the hoisting cable when the statue is at
rest and a) completely immersed; b) out of the water?
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University Physics, Chapter
14
14.3 Buoyancy
Solution:
• Identify : When the statue is immersed, it experiences an upward
buoyant force = in magnitude to the weight of the fluid displaced.
Three forces are acting on the statue: weight, the buoyant force, and
the tension in the cable.
• Set Up : Our target variable is the tension T .
We are given mg, we
can calculate the buoyant force B by using Archimedes’s principle.
• Execute: a) To find B, we first find the volume of the statue, using
the density of gold from table 14.1:
V=
m
ρgold
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=
15.0 kg
19.3 ×103 kg/m3
−
4
3
= 7.77 ×10 m
University Physics, Chapter
14
14.3 Buoyancy
• Using
table 14.1 again, we find the weight of this volume of
seawater:
wsw = msw g = ρ sw V g
= (1.03×103 kg/m3)(7.77 ×10−4 m3)( 9.80 m / s2 )
= 7.84 N
This equals the buoyant force B.
• The statue is at rest, so the net external force acting on it is zero.
∑ Fy = B + T + (−mg ) = 0
T = mg − B = (15.0kg)( 9.80 m / s2 ) − 7.84 N
= 147 N - 7.84 N = 139 N
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University Physics, Chapter
14
14.3 Buoyancy
• If a spring scale is attached to the upper end of the cable, it will
indicate 7.84 N less than if the statue were not immersed in seawater.
• Hence the submerged statue seems to weight 139 N, about 5% less
than its actual weight of 147 N.
b) The density of air is about 1.2 kg/m3, so the buoyant force of air
on the statue is
B = ρair V g = (1.2 kg/m3 )( 7.77 ×10−4 m3 )( 9.80m / s2 )
= 9.1×10−3 N
• Thus the tension in the cable with the statue in air is equal to the
statue’s weight, 147 N.
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University Physics, Chapter
14
14.3 Buoyancy
Example:
The human brain is immersed in cerebrospinal fluid of density 1.007
g/cm3, density is slightly less than the average density of the brain
1.04 g/cm3. Thus most of the weight of the brain is supported by the
buoyant force of the surrounding fluid. What fraction of the weight
of the brain is not supported by this force?
Solution:
• This
problem is an application of the Archimedes principle the
weight of the brain is:
w = ρbrainVbrain g
• The
buoyant force in turn for the brain fully impressed in the
cerebrospinal fluid:
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University Physics, Chapter
14
14.3 Buoyancy
B = ρcerebrospinalVbrain g
• We
know that these two force do not balance each medulla
oblongata to the spinal cord which exerts an additional force on the
brain to determine the fraction of the weight of the brain that is not
balanced by buoyant force we calculate:
w − B ρbrain − ρcerebrospinal 1.04 − 1.007
=
=
= 0.032
w
ρbrain
1.04
• Just
3.2% of the brain’s weight is not balanced by the
cerebrospinal fluid, requiring only a small force to be exerted by the
spinal cord on the brain.
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University Physics, Chapter
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14.3 Buoyancy
Surface Tension
• The surface of the liquid behaves like a membrane under tension.
• Surface tension arises because the molecules of the liquid exert
attractive forces on each other.
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University Physics, Chapter
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14.3 Buoyancy
Surface Tension
• There
is zero net force on a
molecule inside the volume of the
liquid.
• But
a surface molecule is
drawn into the volume (figure
below). Thus the liquid tends to
minimize its surface area, just as
a stretched membrane does.
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University Physics, Chapter
14
14.3 Buoyancy
Surface Tension
• Surface tension explains why freely falling raindrops are spherical
(not teardrop-shaped): a sphere has a smaller surface area for its
volume than any other shape.
• It also explains why hot, soapy water is used for washing. To wash
clothing thoroughly, water must be forced through the tiny spaces
between the fibers.
• To do so requires increasing the surface area of the water, which is
difficult to achieve because of surface tension.
• The job is made easier by increasing the temperature of the water
and adding soap, both of which decrease the surface tension.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.3 Buoyancy
Surface Tension
• Surface
tension is important for a millimeter-sized water drop,
which has a relatively large surface area for its volume.
• For large quantities of liquid, the ratio of surface area to volume is
relatively small, and surface tension is negligible compared to
pressure forces.
• For the remainder of this chapter, we will consider only fluids in
bulk and hence will ignore the effects of surface tension.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
• We
will now consider motion of a fluid. Fluid flow can be
extremely complex, as shown by the currents in river rapids.
• An ideal fluid is a fluid that is incompressible (that is, its density
cant change) and has no internal frication (called viscosity).
• The path of an individual particle in a moving fluid is called a flow
line.
• If the overall flow pattern does not change with time, the flow is
called steady flow.
•A
streamline is a curve whose tangent at any point is in the
direction of the fluid velocity at that point.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
• We will consider only steady-flow situations, for which flow lines
and streamlines are identical.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
• The flow lines passing through the edge of an imaginary element of
area A , form a tube called a flow tube (see figure 14.19/466).
• From the definition of a flow line, in steady flow no fluid can cross
the side walls of a flow tube; the fluids in different flow tubes cant
mix.
• Laminar flow, in which adjacent layers of fluid slide smoothly past
each other and the flow is steady.
• At sufficiently high flow rates, or when
boundary surface cause
abrupt changes in velocity, the flow can become irregular and
chaotic. This is called turbulent flow.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
Laminar flow
Dr. Y. Abou-Ali, IUST
Turbulent flow
University Physics, Chapter
14
14.4 Fluid Flow
The Continuity Equation
• The mass of a moving fluid does not change as it flows. This leads
to an important quantitative relationship called the continuity
equation.
v2
dV2 = A2v2dt
dV1 = A1v1dt
Dr. Y. Abou-Ali, IUST
v1
University Physics, Chapter
14
14.4 Fluid Flow
The Continuity Equation
• Let
us consider the case of an incompressible fluid so that the
density ρ has the same value at all point.
• The mass dm 1 flowing into the tube across A 1 in time dt is:
dm1 = ρ A1v1 dt
• The mass dm 2 flowing into the tube across A 2 in time dt is:
dm 2 = ρ A2 v 2 dt
• In steady flow the total mass in the tube is constant, so dm 1 = dm 2,
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
The Continuity Equation
ρ A1 v1 d t = ρ A 2 v 2 d t
A1 v 1 = A 2 v 2
(1 4 .1 0 )
( c o n tin u ity e q u a tio n , in c o m p r e s s ib le f lu id )
• The product Av
is the volume flow rate dV /dt , the rate at which
volume crosses a section of the tube:
dV
= Av
( v o lu m e flo w ra te )
(1 4 .1 1 )
dt
• The mass flow rate is the mass flow per unit time through a cross
section. This is equal to density ρ times the volume flow rate dV /dt .
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
The Continuity Equation
• We can generalise Eq. (14.10):
ρ1 A1v1 = ρ 2 A2 v2
(continuity equation, compressible fluid) (14.12)
Example 14.6 ( Incompressible fluid flow):
As part of a lubricating system for heavy machinery, oil of density
850 kg/m3 is pumped through a cylindrical pipe of diameter 8.0 cm
at a rate of 9.5 liters per second. a) What is the speed of the oil?
What is the mass flow rate? b) If the pipe diameter is reduced to 4.0
cm, what are the new values of the speed and volume flow rate?
Assume that the oil is incompressible.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
Solution:
• Identify and Set Up : we use Eq. (14.11), to determine the speed v 1
in the 8.0 cm-diameter section. Using Eq. (14.10), to find the speed v 2
in the 4.0 cm-diameter section.
• Execute: a) The volume flow rate dV /dt equals the product A 1v 1,
where A 1 is the cross-sectional area of the pipe of diameter 8.0 cm
and radius 4.0 cm. Hence:
dV / dt
v1 =
=
A1
−
3
3
( 9.5 L/s)( 10 m /L)
π (4.0 ×10−2 m)2
= 1.9 m/s
The mass flow rate is: ρ dV /dt = (850 kg/m3)(9.5 × 10-3 m3/s) = 8.1
kg/s.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.4 Fluid Flow
b) Since the oil is incompressible, the volume flow rate has the same
value (9.5 L/s) in both sections of pipe. From Eq. (14.10),
A1
π (4.0 ×10−2 m)2
v2 =
v1 =
(1.9 m/s) = 7.6 m/s
−
2
2
A2
π (2.0 ×10 m)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Speed of Blood in Capillaries
Example:
The heart of the standard man pumps 5 liters of blood per minute into the aorta:
A) What is the volume flow rate in the cardiovascular system
B) What is the speed of the blood in the aorta
C) If we assume that the blood passes through all systemic capillaries in our body
in series, how fast would it have to flow through each capillary
D) What is the speed of the blood in parallel through the system capillaries in the
human body.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Speed of Blood in Capillaries
Solution:
A) The amount of the blood flowing:
m3/s
B) A typical inner diameter of the aorta is d aorta= 2.2 cm this leads to
cross-section area:
m2
The speed of the blood in aorta is obtained from it’s inner crosssectional area and the volume passing per second using:
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Speed of Blood in Capillaries
We obtain:
This is the frequently used results: blood flows through the aorta at
an average speed of about 20 cm/sec.
C) Let’s assume that blood passes through each single systemic
capillary at the rate found in part (a). For the outer diameter a
capillary we use 9 µm this value leads to an inner diameter of
d capillary = 7 µm the cross section area is:
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Speed of Blood in Capillaries
We use the equation of the continuity to derive the speed for blood:
Then it would no longer be possible to exchange oxygen and
nutrients with the surrounding tissue. The physiological purpose of
the cardiovascular system would be lost.
D) A slow flow of blood in the systemic capillaries is achieved by
arranging them parallel to each other with a combined cross-section
that is large than the cross- section of the aorta. The equation of
continuity allows us to determine the actual speed of blood in the
capillaries once we have know the compined cross-section and the is
2100\cm2 so:
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
Speed of Blood in Capillaries
vcapillary
 ∆V 


 ∆t  aorta
=
= 8.3 ×10 −5 / 0.21m 2 = 4 ×10 − 4 m / s
Acapillary
Again this is a frequently used value blood flows very slowly
through the capillaries at less than 1 mm/s.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Bernoulli’s Equation
• We
can derive an important
relationship
called
Bernoulli's
equation that relates the pressure,
flow speed, and height for flow of an
ideal, incompressible fluid.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Bernoulli’s Equation
• To derive Bernoulli’s equation, we apply the work-energy theorem
to the fluid in a section of a flow tube.
2
2
1
1
p1 + ρ gy1 + ρ v1 = p 2 + ρ gy 2 + ρ v 2
2
2
(Bernoulli’s equation)
(14.17)
• The subscripts 1 & 2 refer to any two points along the flow tube, so
we can also write:
2
1
p + ρ gy + ρ v = constant
2
Dr. Y. Abou-Ali, IUST
(14.18)
University Physics, Chapter
14
14.5 Bernoulli’s Equation
• Blood flow characteristics are changed dramatically by plaque.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Bernoulli’s Equation
Example 14.7 ( Water pressure in the home):
Water enters a house through a pipe with an inside diameter of 2.0
cm at an absolute pressure of 4.0 × 105 Pa (about 4 atm). A 1.0 cm
diameter pipe leads to the second-floor bathroom 5.0 m above (figure
14.24). When the flow speed at the inlet pipe is 1.5 m/s, find the flow
speed, pressure, and volume flow rate in the bathroom.
Solution:
• Identify and Set Up : Let points 1 & 2 be at the inlet pipe and at the
bathroom, respectively. We take y 1 = 0 (at the inlet) and y 2 = 5.0 m
(at the bathroom). Our first two targets are the speed v 1 and
pressure p 1.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Bernoulli’s Equation
• Execute: From continuity equation, Eq. (14.10):
A1
π (1.0 cm) 2
(1.5 m / s) = 6.0 m / s
v2 =
v1 =
A2
π (0.50 cm) 2
We are given p 1 and v 1, and can find p 2 from Bernoulli’s equation:
p2 = p1 − 1 ρ (v22 − v12 ) − ρ g ( y2 − y1) = 4.0 ×105 Pa
2
− 1 (1.0 ×103 kg / m3 )( 36 m2 / s2 − 2.25 m2 / s2 )
2
− (1.0 ×103 kg / m3 )( 9.8 m / s2 )( 5.0 m)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Bernoulli’s Equation
5
5
5
= 4.0 ×10 Pa - 0.17 ×10 Pa - 0.49 ×10 Pa
= 3.3 ×105 Pa = 3.3 atm
The volume flow rate is:
dV
= A2v2 = π (0.50 ×10−2 m2 )2 (6.0 m / s)
dt
= 4.7 ×10−4 m3 / s = 0.47 L / s
• Read Example: 14.8/470 & 14.9/471.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.5 Viscosity and Turbulence
Viscosity
• Viscosity is internal friction in a fluid. Viscous forces oppose the
motion of one portion of a fluid relative to another.
• Viscosity
is the reason why it takes effort to paddle a canoe
through calm water.
• Viscous effects are important in the flow of fluids in pipes, the flow
of blood, and the lubrication of engine parts.
• Fluids that flow readily, such as water or gasoline, have smaller
viscosities than do “thick” liquids such as honey or motor oil.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.6 Viscosity and Turbulence
Turbulence
• When the speed of a flowing fluid exceeds a certain critical value,
the flow is no longer laminar.
• Instead,
the flow pattern becomes extremely irregular and
complex, and it changes continuously with time; there is no steadystate pattern..
• This irregular, chaotic flow is called turbulence.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14
14.6 Viscosity and Turbulence
Dr. Y. Abou-Ali, IUST
University Physics, Chapter
14