Download Chapter 9: Fluids

Chapter 9: Fluids
•Introduction to Fluids
•Pressure
•Pascal’s Principle
•Gravity and Fluid Pressure
•Measurement of Pressure
•Archimedes’ Principle
•Continuity Equation
•Bernoulli’s Equation
•Viscosity and Viscous Drag
•Surface
S rface Tension
1
Fluids
A liquid will flow to take the shape of the container that holds
it A gas will completely fill its container
it.
container.
Fluids are easily deformable by external forces.
A liquid is incompressible. Its volume is fixed and is
impossible to change
change.
2
FLUIDS (both liquids and gases)Best quantity to characterize a fluid is the density ρ
ρ = m/V [kg/m³]
m=Mass,
M
V=
V Volume
V l
ρwater= 1000 kg/m³
Mass of 1L H2O (L = liter = 1dm³)
m1L Water = 1000 kg/m³ 1dm³ = 1000 kg/m³ 10-³m³ = 1kg
1 gallon of water = 3.79 L = 3.79kg
Substance
Density (kg/m3)
Substance
Density (kg/m3)
Gold
19,300
Olive oil
920
Mercury
13,600
Ice
917
Lead
11 300
11,300
Ethyl alcohol
806
Silver
10,500
Cherry (wood)
800
Iron
7860
Balsa wood
120
Aluminum
2700
Styrofoam
100
Ebony (wood)
1220
Oxygen
1.43
Whole blood (37 ºC)
1060
Air
1.29
Sea ater
Seawater
1025
Heli m
Helium
0 179
0.179
Fresh water
1000
3
3
Volumerefr = 1 ⋅ 0.6 ⋅ 0.75m = 0.45m
3
G
G
Weight Air , refr = (0.45m ⋅1.29 3 ) ⋅ g = (580.5 g ) ⋅ g
m
G
G
Weighteggs , refr = (12 ⋅ 44 g ) ⋅ g = (528 g ) ⋅ g
3
kg
4
Static
Dynamic
5
Pressure
Pressure arises from the collisions between the particles of
a fluid with another object (container walls for example).
There is a momentum
change (impulse) that is
away from the container
walls. There must be a
force exerted on the
particle by the wall.
6
By Newton’s 3rd Law, there is a force on the wall due to the particle.
Pressure is defined as
F
P= .
A
The units of pressure are N/m2 and are called Pascals (Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
(older unit: 1 bar = 105 Pa)
Are we under pressure for our entire life? - YES
Air above us has a weight.
weight
Atmospheric Pressure Pat= 1.01 x 105 N/m² = 101 kPa ≈ 1 bar
7
(average value on earth´s surface, the higher you go the lower the pressure)
pascal
(Pa)
bar
(bar)
technical
atmosphere
(at)
atmosphere
(atm)
torr
(Torr)
pound-force per
square inch
(psi)
1 Pa
≡ 1 N/m2
10−5
1.0197×10−5
9.8692×10−6
7.5006×10−3
145.04×10−6
1 bar
100,000
≡ 106 dyn/cm2
1.0197
0.98692
750.06
14.5037744
1 at
98,066.5
0.980665
≡ 1 kgf/cm2
0.96784
735.56
14.223
1 atm
101,325
1.01325
1.0332
≡ 1 atm
760
14.696
1 torr
133.322
1.3332×10−3
1.3595×10−3
1.3158×10−3
≡ 1 Torr;
≈ 1 mmHg
19.337×10−3
1 psi
6,894.76
68.948×10−3
70.307×10−3
68.046×10−3
51.715
≡ 1 lbf/in2
8
9
The pressure in a fluid (gas, liquid) acts equally in all
directions and acts at right angles to any surface
10
Example (text problem 9.1): Someone steps on your toe,
exerting a force of 500 N on an area of 1
1.0
0 cm2. What is the
average pressure on that area in atmospheres?
2
⎞
2⎛ 1m
−4
2
1.0 cm ⎜
⎟ = 1.0 ×10 m
⎝ 100 cm ⎠
F
500 N
Pav = =
A 1.0 ×10 -4 m 2
1 atm
⎞
6
2 ⎛ 1 Pa ⎞⎛
= 5.0 × 10 N/m ⎜
⎟
2 ⎟⎜
5
⎝ 1 N/m ⎠⎝ 1.013 × 10 Pa ⎠
= 49 atm
11
Gravity’s
Gravity
s Effect on Fluid Pressure
(What you feel at a start/landing of an airplane or when you go
scuba diving)
FBD for the fluid cylinder
P1A
A cylinder
of fluid
fl id
w
P2A
12
Apply
A
l N
Newton’s
t ’ 2ndd Law
L
to the fluid cylinder:
∑F = P A− PA− w = 0
2
1
P2 A − P1 A − (ρAd )g = 0
P2 − P1 − ρgd = 0
∴ P2 − P1 = ρgd
or P2 = P1 + ρgd
If P1 (the pressure at the top of the cylinder) is known, then
th above
the
b
expression
i can b
be used
d tto fifind
d th
the variation
i ti off
pressure with depth in a fluid.
13
If the top of the fluid column is placed at the surface of the
fluid then P1=P
fluid,
Patm
open.
t if the container is open
P = Patm + ρgd
Example (text problem 9.15): At the surface of a
freshwater lake
lake, the pressure is 105 kPa.
kPa (a) What is the
pressure increase in going 35.0 m below the surface?
P = Patm + ρgd
d
ΔP = P − Patm = ρgd
35m
(
)(
)
= 1000 kg/m 9.8 m/s (35 m )
3
2
= 343 kPa = 3.4
3 4 atm
14
Example: The surface pressure on the planet Venus is 95
atm.
t
How
H
f below
far
b l
th
the surface
f
off the
th ocean on Earth
E th do
d
you need to be to experience the same pressure? The
densityy of seawater is 1025 kg/m
g 3.
P = Patm + ρgd
95 atm = 1 atm + ρgd
ρgd = 94 atm = 9.5 × 10 N/m
6
(1025 kg/m )(9.8 m/s )d = 9.5
9 5 × 10
3
2
6
N/m
2
2
d = 950 m ≈ 0.600miles
15
16
17
Measuring Pressure
A manometer
i aU
is
U-shaped
h
d
tube that is
partially
p
y filled
with liquid.
Both ends of the
tube are open to
the atmosphere.
P 1 = P2
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A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the
p
a force will be exerted on the fluid in the U-tube.
atmosphere,
This changes the equilibrium position of the fluid in the tube.
19
From the figure:
At point C
Also
Pc = Patm
PB = PB'
The pressure at point B is the pressure of the gas.
PB = PB ' = PC + ρgd
PB − Patm = ρgd
Pgauge = ρgd
Gauge pressure is the pressure relative to the local
atmospheric or ambient pressure.
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A Barometer
P2: pressure in glass tube
P1 = P2 + ρhg = 0 + ρhg
Therefore:
Pat= ρhg
1 atmosphere =
Pat = 760 mm Hg
The atmosphere pushes on the container of mercury which
forces mercury up the closed, inverted tube. The distance
d is called the barometric pressure.
21
Example (text problem 9.22): An IV is connected to a
patient’s
patient
s vein.
vein The blood pressure in the vein has a gauge
pressure of 12 mm of mercury. At least how far above the
vein must the IV bag be placed in order for fluid to flow into
the vein? Assume that the density of the IV fluid is the same
as blood.
Blood: Pgauge =
ρ Hg gh1
h1 = 12 mm
IV:
The pressure is
Th
i equivalent
i l t
to raising a column of
mercuryy 12 mm tall.
Pgauge = ρ blood gh2
22
At a minimum, the gauge pressures must be equal. When
h2 is large enough, fluid will flow from high pressure to low
pressure.
Pgauge = ρ Hg gh1 = ρ blood gh2
ρ Hgg gh1
h2 =
ρ blood g
⎛ ρ Hg
⎞
⎛ 13,600 kg/m 3 ⎞
H
⎟⎟h1 = ⎜⎜
⎟(12 mm )
= ⎜⎜
3 ⎟
⎝ 1060 kg/m ⎠
⎝ ρ blood ⎠
= 154 mm
23
Pascal’s
Pascal
s Principle
A change in pressure at any point in a confined fluid is
transmitted everywhere throughout the fluid
fluid. (This is useful
in making a hydraulic lift.)
The applied force is
transmitted to the
piston of crosssectional area A2 here.
here
Apply a force F1 here
to a piston of crosssectional area A1.
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Mathematically,
ΔP at point 1 = ΔP at point 2
F1
F2
=
A1 A 2
⎛ A2 ⎞
⎟⎟ F1
F2 = ⎜⎜
⎝ A1 ⎠
25
Example: Assume that a force of 500 N (about 110 lbs) is
applied
li d tto th
the smaller
ll piston
i t iin th
the previous
i
fifigure. F
For each
h
case, compute the force on the larger piston if the ratio of the
piston areas ((A2/A1) are 1, 10, and 100.
p
Using
g Pascal’s Principle:
A2 A1
1
10
100
F2
500 N
5000 N
50,000 N
26
The work done pressing the smaller piston (#1) equals the
work done by the larger piston (#2).
W1 = W2
F1d1 = F2 d 2
F1
d2 =
d1
F2
27
Example: In the previous example, for the case A2/A1 = 10,
it was ffound
d th
thatt F2/F1 = 10.
10 If the
th larger
l
piston
i t needs
d tto
rise by 1 m, how far must the smaller piston be depressed?
Using the result on the previous slide,
F2
d1 =
d 2 = 10 m
F1
28
Archimedes’ Principle
An object completely immersed in a fluid experiences an
upward buoyant force equal in magnitude to the weight of
fluid displaced by the object.
F1
An FBD for an object
floating submerged in
a fluid.
w
F2
The total force on the block due to the fluid is called the
buoyant force.
FB = F2 − F1
where F2 > F1
29
The magnitude of the buoyant force is:
FB = F2 − F1
= P2 A − P1 A
= (P2 − P1 )A
From before:
P2 − P1 = ρg
gd
The result is
FB = ρgdA = ρ fluid gV fluid ,displaced
30
Archimedes’ Principle:
p A fluid exerts an upward
p
buoyant
y
force on a submerged object equal in magnitude to the
weight of the volume of fluid displaced by the object.
FB = ρ fluid gV fluid ,displaced
31
32
33
Example (text problem 9.30): A flat-bottomed barge loaded
with coal has a mass of 3
3.0×10
0×105 kg.
kg The barge is 20
20.0
0 m long
and 10.0 m wide. It floats in fresh water. What is the depth of
the barge below the waterline?
∑F = F
B
FB
FBD
for the
barge
−w=0
FB = w
mw g = (ρ wVw )g = mb g
w
ρ wVw = mb
ρ w ( Ad ) = mb
mb
3.0 ×105 kg
d=
=
= 1.5 m
3
ρ w A 1000 kg/m (20.0 m *10.0 m )
(
)
34
35
Example (text problem 9.34): A piece of metal is released
under water.
water The volume of the metal is 50
50.0
0 cm3 and its
specific gravity is 5.0. What is its initial acceleration? (Note:
when v=0, there is no drag force.)
FB
FBD
for the
metal
w
∑F = F
B
− w = ma
The buoyant force is the
weight
g of the fluid displaced
p
by the object
FB = ρ waterVgg
⎛ ρ waterV
⎞
ρ waterVg
FB
−g =
− g = g⎜
− 1⎟
Solve for a: a =
⎜ρ V
⎟
m
ρ objectVobject
⎝ object object ⎠
36
Example continued:
Since the object is completely submerged V=Vobject.
specific
ifi gravity
it =
ρ
ρ water
3 is the
where ρwater
=
1000
kg/m
t
density of water at 4 °C.
ρ object
= 5.0
Given specific gravity =
ρ water
⎛ ρ waterV
⎞
⎛ 1
⎞
⎛ 1
⎞
⎜
⎟
a=g
−1 = g⎜
− 1⎟ = g ⎜
− 1⎟ = −7.8 m/s 2
⎜ρ V
⎟
⎝ S .G. ⎠
⎝ 5.0 ⎠
object
object
⎝
⎠
37
38