Download Lecture 13

Physics 106 Lecture 13
Fluid Mechanics
SJ 7th Ed.: Chap 14.1 to 14.5
•
•
•
•
•
•
•
•
•
•
•
What is a fluid?
Pressure
Pressure varies with depth
Pascal’s principle
Methods for measuring pressure
Buoyant forces
Archimedes principle
y
assumptions
p
Fluid dynamics
An ideal fluid
Continuity Equation
Bernoulli’s Equation
What is a fluid?
Solids: strong intermolecular forces
ƒ
ƒ
ƒ
definite volume and shape
rigid crystal lattices, as if atoms on stiff springs
deforms elastically (strain) due to moderate stress
(pressure) in any direction
Fluids: substances that can “flow”
ƒ
ƒ
ƒ
no definite shape
molecules are randomly arranged, held by weak cohesive intermolecular
forces and by the walls of a container
liquids and gases are both fluids
Liquids: definite volume but no definite shape
ƒ
ƒ
ƒ
often almost incompressible under pressure (from all sides)
can not resist tension or shearing (crosswise) stress
no long range ordering but near neighbor molecules can be held weakly together
Gases: neither volume nor shape are fixed
ƒ
molecules move independently of each other
ƒ
comparatively easy to compress: density depends on temperature and pressure
Fluid Statics - fluids at rest (mechanical equilibrium)
Fluid Dynamics – fluid flow (continuity, energy conservation)
1
Mass and Density
• Density is mass per unit volume at a point:
ρ≡
•
•
Δm
ΔV
or
ρ≡
m
V
• scalar
• units are kg/m3, gm/cm3..
• ρwater= 1000 kg/m3= 1.0 gm/cm3
Volume and density vary with temperature - slightly in liquids
The average molecular spacing in gases is much greater than in liquids.
Force & Pressure
• The pressure P on a “small” area ΔA is the ratio of the
magnitude of the net force to the area
P=
•
•
ΔF
or P = F / A
ΔA
G
G
ΔF = PΔA = PΔAn̂
n̂
PΔA n
Pressure is a scalar while force is a vector
The direction of the force producing a pressure is perpendicular
to some area of interest
•
h
At a point in a fluid (in mechanical
equilibrium) the pressure is the same in
any direction
Pressure units:
ƒ 1 Pascal (Pa) = 1 Newton/m2 (SI)
ƒ 1 PSI (Pound/sq. in) = 6894 Pa.
ƒ 1 milli-bar = 100 Pa.
2
Forces/Stresses in Fluids
•
Fluids do not allow shearing stresses or tensile stresses (pressures)
Tension
Shear
Compression
The only stress that can be exerted on an object submerged in a
static fluid is one that tends to compress the object from all sides
The force exerted by a static fluid on an object is always
perpendicular to the surfaces of the object
•
•
Question: Why can you push a pin easily into a potato, say, using
very little force, but your finger alone can not push into the skin even
if you push very hard?
A sensor to measure absolute pressure
•
•
•
•
The spring is calibrated by a known force
Vacuum is inside
The force due to fluid pressure presses on the top of the piston and
compresses the spring until the spring force and F are equal.
The force the fluid exerts on the piston is then measured
3
Pressure in a fluid varies with depth
Fluid is in static equilibrium
The net force on the shaded volume = 0
•
Incompressible liquid - constant density ρ
•
•
Horizontal surface areas = A
Forces on the shaded region:
y=0
y1
– Upward force on bottom:
∑ Fy = 0 =
•
P1
F2
– Weight of shaded fluid: Mg
– Downward force on top:
F1
h
y2
P2
Mg
F1 =P1A
F2 = P2A
P2 A − P1 A − Mg
In terms of density, the mass of the
shaded fluid is:
The extra pressure at
extra depth h is:
M = ρΔV = ρAh
ΔP = P2 − P1 = ρgh
∴ P2 A ≡ P1 A + ρghA
h ≡ y1 − y 2
Pressure relative to the surface of a liquid
air
Example: The pressure at depth h is:
P0
Ph = P0 + ρgh
ƒ
ƒ
ƒ
ƒ
ƒ
P0 is the local atmospheric (or ambient)
pressure
Ph is the absolute pressure at depth h
The difference is called the gauge pressure
All points at the same depth are at the
same pressure; otherwise, the fluid could
not be in equilibrium
The pressure at depth h does not depend
on the shape of the container holding the
fluid
P0
h
Ph
liquid
Preceding
P
di equations
ti
also
l
hold approximately for
gases such as air if the
density does not vary
much across h
Ph
4
Atmospheric pressure and units conversions
• P0 is the atmospheric pressure if the liquid is open to the
atmosphere.
• Atmospheric pressure varies locally due to altitude,
temperature, motion of air masses, other factors.
• Sea level atmospheric pressure P0 = 1.00 atm
= 1.01325 x 105 Pa = 101.325 kPa = 1013.25 mb (millibars)
= 29.9213” Hg = 760.00 mmHg ~ 760.00 Torr
= 14.696 psi (pounds per square inch)
Pascal
(Pa)
1 Pa
bar (bar)
2
atmosphere
(atm)
−5
≡ 1 N/m
−6
10
9 8692×10
9.8692×10
torr
(Torr)
pound-force per
square inch (psi)
−3
7 5006×10
7.5006×10
−6
145 04×10
145.04×10
1 bar
100,000
≡
106 dyn/cm2
0.98692
750.06
14.5037744
1 atm
101,325
1.01325
≡ 1 atm
760
14.696
1 torr
133.322
1.3332×10−3
1.3158×10−3
≡ 1 Torr;
≈ 1 mmHg
19.337×10−3
1 psi
6.894×103
68.948×10−3
68.046×10−3
51.715
≡ 1 lbf/in2
Pressure Measurement: Barometer
•
•
near-vacuum
•
•
Invented by Torricelli (1608-47)
Measures atmospheric pressure P0 as it varies
with the weather
The closed end is nearly a vacuum (P = 0)
One standard atm = 1.013 x 105 Pa.
P0 = ρ Hggh
Mercury (Hg)
How high is the Mercury column?
P
1.013 × 10 5 Pa
h = 0 =
= 0.760 m
ρ Hgg (13.6 × 10 3 kg / m3 )(9.80 m/s 2 )
One 1 atm = 760 mm of Hg
= 29.92 inches of Hg
How high would a water column be?
h =
P0
ρ water g
=
1.013 × 10 5 Pa
(1.0 × 103 kg / m3 )(9.80 m/s 2 )
= 10.34 m
Height limit for a suction pump
5
Pressure Measurements: Manometer
Measures the pressure of the gas in the
closed vessel
One end of the U-shaped tube is open to
the atmosphere at pressure P0
The other end is connected to the
pressure PA to be measured
Points A and B are at the same elevation.
The pressure PA supports a liquid
column of height h
The Pressure of the gas is:
•
•
•
•
•
•
PA = PB = P0 + ρgh
Absolute vs. Gauge Pressure
What effect would
increasing P0 have on
the column height?
The gauge pressure is PA – P0= ρgh
– What you measure in your tires
Gauge pressure can be positive or negative.
PA is the absolute pressure – you have to add the ambient pressure P0 to
the gauge pressure to find it.
•
•
•
Example: Blood Pressure
•
•
•
Normal Blood pressure is: 120 (systolic = peak)
over 70 (diastolic = resting) – in mmHg.
What do these numbers mean?
How much pressure do they correspond to in Pa. and psi?
Recall: 760.00 mmHg = 1.01325 x 105 Pa =14.696 psi
(101,325/760.00) = 133.322 Pa./mmHg
(101,325/14.696) = 6894. Pa./psi
Systolic pressure of 120
= 16,000 Pa.
= 2
2.32
32 psi
Diastolic pressure of 70
= 9333 Pa.
= 1.354 psi
Mercury
Sphygmomanometer
Aneroid
Sphygmomanometer
Gauge or absolute?
6
Pascal’s Principle
A change in the pressure applied to an enclosed incompressible fluid
is transmitted undiminished to every point of the fluid and to the
walls of the container.
ΔF
ΔP
Example: open container
The pressure in a fluid depends on depth h and on the
value of P0 at the surface
All points at the same depth have the same pressure.
•
p0
•
ph
Ph = P0 + ρgh
•
Add piston
i t
off area A with
ith lead
l d balls
b ll on it & weight
i ht W.
W
Pressure at surface increases by ΔP = W/A
Pext = P0 + ΔP
•
Ph
Pressure at every other point in the fluid (Pascal’s law),
increases by the same amount, including all locations at
depth h.
Ph = Pext + ρgh
Pascal’s Law Device - Hydraulic press
A small input force generates a large output force
•
•
Assume the working fluid is incompressible
Neglect the (small here) effect of height on pressure
The volume of liquid pushed down on
th left
the
l ft equals
l the
th volume
l
pushed
h d up
on the right, so:
•
A1 Δx 1 = A 2 Δx 2
A
Δx 2
∴
= 1
Δx 1
A2
•
Other hydraulic lever devices
using Pascal’s Law:
• Squeezing a toothpaste tube
• Hydraulic brakes
• Hydraulic jacks
• Forklifts, backhoes
Assume no loss of energy in
the fluid
fluid, no friction
friction, etc.
etc
Work1 = F1Δx1 = Work 2 = F2 Δx 2
mechanical advantage
∴
F2 Δx 1
A
=
= 2
F1 Δx 2
A1
7
Archimedes Principle
•
•
•
•
C. 287 – 212 BC
Greek mathematician, physicist and engineer
Computed π and volumes of solids
Inventor of catapults, levers, screws, etc.
Discovered nature of buoyant force – Eureka!
Why do ships float and sometimes sink?
Why do objects weigh less when submerged in a fluid?
identical pressures at every point
ball of liquid
in equilibrium
•
hollow ball
same
upward force
An object immersed in a fluid feels an upward buoyant force that equals
the weight of the fluid displaced by the object. Archimedes’s Principle
– The fluid pressure increases with depth and exerts forces that are the same
whether the submerged object is there or not.
– Buoyant forces do not depend on the composition of submerged objects.
– Buoyant forces depend on the density of the liquid and g.
Archimedes’s principle - submerged cube
A cube that may be hollow or made of some material is submerged in a fluid
Does it float up or sink down in the liquid?
•
The extra pressure at the lower surface
compared to the top is:
ΔP = Pbot
= ρ flgh
b t − Ptop
t
•
•
•
The pressure at the top of the cube causes
a downward force of Ptop A
The larger pressure at the bottom of the
cube causes an upward force of Pbot A
The upward buoyant force B is the weight
of the fluid displaced by the cube:
B = ΔPA = ρ flghA = ρ flgV = Mflg
•
The weight of the actual cube is:
Fg = Mcubeg = ρcubegV
The cube rises if B > Fg (ρfl > ρcube)
The cube sinks if Fg> B (ρcube > ρfl)
Similarly for irregularly shaped
objects
ρcube is the average density
8
Archimedes's Principle: totally submerged object
Object – any shape - is totally submerged in a fluid of density ρfluid
The upward buoyant force is the
weight of displaced fluid:
B = ρ fluidgVfluid
The downward gravitational force on the
object is:
Fg = Mobject g = ρobject gVobject
The volume of fluid displaced and the
object’s volume are equal for a totally
submerged object. The net force is:
Fnet = B − Fg = [ρ fluid − ρobject ] g Vobject
The direction of the motion of an object
in a fluid is determined only by the
densities of the fluid and the object
–
–
If the density of the object
the density of the fluid, the
object accelerates upward
If the density of the object
the density of the fluid, the
object sinks
is less than
unsupported
The apparent weight is the
external force needed to
restore equilibrium, i.e.
is more than
unsupported
Wapparent = Fg − B = − Fnet
Archimedes’s Principle: floating object
An object sinks or rises in the fluid until it reaches equilibrium.
The fluid displaced is a fraction of the object’s volume.
At equilibrium the upward buoyant
force is balanced by the downward
weight of the object:
Fnet = B − Fg = 0 ⇒ B = Fg
The volume of fluid displaced Vfluid corresponds to
the portion of the object’s volume below the fluid
level and is always less than the object’s volume.
Equate:
B = ρ fluidgVfluid = Fg = ρobjectgVobject
Solving:
ρ fluid Vfluid = ρobject Vobject
ρobject
Vfluid
=
ρ fluid
Vobject
‰ Objects float when their average
density is less than the density of
the fluid they are in.
‰ The ratio of densities equals the
fraction of the object’s volume that
is below the surface
9
Archimedes’s Principle: King’s Crown
King to Archimedes: “Is the crown made of pure gold?”
• The crown’s density ρcrown = mcrown/ Vcrown should = ρgold = 19,300 kg/m3
• Weigh the crown in air to find mcrown
• But: how to find Vcrown??
Solution:
• Weigh crown in air to find mcrown
• Weigh it again fully submerged in a fluid
• The apparent weight T2 is:
T2 = Fg − B
•
The Buoyant force equals the
difference in scale readings...
•
solve:
B = Fg − T2 = ρ flgVcrown
Vcrown =
Fg − T2
ρ flg
∴ ρcrown =
Fg
Fg − T2
ρ fl
Example: What fraction of an iceberg is underwater?
Apply Archimedes’ Principle
displaced seawater
ρobject
Vunderwater
Vfluid
=
=
Vtotal
Vobject
ρ fluid
glacial fresh water ice
From table:
ρobject = ρice = 0.917 × 10 3 kg/m3
ρ fluid = ρ seawater = 1.03 × 10 3 kg/m3
Water expands when it freezes. If not....
...ponds, lakes, seas freeze to the bottom in winter
∴
Vunderwater
= 89.03%
Vtotal
t t l
What if iceberg is in a freshwater lake?
ρ fluid = ρ freshwater = 1.00 × 10 3 kg/m3
Vunderwater
= 91.7%
Vtotal
Floating objects are more buoyant in saltwater
Freshwater tends to float on top of seawater...
10
Water has special properties
Maximum density at 4o C.
• ρwater= 1000 kg/m3= 1.0 gm/cm3
• turnover of ponds and lakes
• ice floats, so bodies of water
do not freeze to the bottom
Fluids’ Flow is affected by their viscosity
•
Viscosity measures the internal friction in a fluid.
Low viscosity
• gases
•
•
Medium viscosity
• water
• other fluids that pour
and flow easily
y
High viscosity
• honey
• oil and grease
•g
glass
Viscous forces depend on the resistance that two adjacent
layers of fluid have to relative motion.
Part of the kinetic energy of a fluid is converted to internal
energy, analogous to friction for sliding surfaces
Ideal Fluids – four approximations to simplify the
analysis of fluid flow:
•
•
•
•
The fluid
Th
fl id iis nonviscous
i
– internal
i
l f
friction
i i iis neglected
l
d
The flow is laminar (steady, streamline flow) – all particles
passing through a point have the same velocity at any time.
The fluid is incompressible – the density remains constant
The flow is irrotational – the fluid has no angular momentum
about any point. A small paddle wheel placed anywhere does
not feel a torque and rotate
11
Laminar Flow - Streamlines
Steady state flow – at any point the velocity of particles is constant
Each particle of the fluid follows a smooth path called a streamline.
Particles’ velocities are tangent to streamlines (field lines).
The paths of different particles never cross each other
Every given fluid particle arriving at a given point has the same velocity
ƒ
ƒ
ƒ
ƒ
ƒ
velocity field lines
Turbulent flow
ƒ
ƒ
ƒ
An irregular flow characterized by small whirlpool-like regions
Turbulent flow occurs when the particles go above some critical
(Mach) speed
Energy is dissipated, more work must be done to maintain flow
12
Flow of an ideal fluid through a short section of pipe
Constant density and velocity within volume element dV
Incompressible fluid means dρ/dt = 0
Mass flow rate = amount of mass crossing area A per unit time
= a “current”sometimes called a “mass flux”
cross-section area A
velocity v
volume of fluid in cylinder
Imass = mass flow rate =
= ρA
length dx
dx
= ρAv
dt
= dV = Adx
dM d
=
(ρV )
dt dt
Imass ≡ mass flow rate = ρAv
Ivol ≡ volume flow rate = Av
Jmass ≡ mass flow/unit area = ρv
Equation of Continuity: conservation of mass
•
•
•
•
An ideal fluid is moving through a pipe of nonuniform diameter
The particles move along streamlines in steady-state flow
The mass entering at point 1 cannot disappear or collect in the pipe
The mass that crosses A1 in some time interval is the same as the mass that
crosses A2 in the same time interval.
mass flow in =
•
ρ1 A1v1 = mass flow out = ρ 2 A 2 v 2
The fluid is incompressible so:
ρ2
ρ1 = ρ 2 = a cons tan t
∴ A 1 v1 = A 2 v 2
•
•
This is called the equation of continuity
for an incompressible
p
fluid
The product of the area and the fluid
speed (volume flux) at all points along a
pipe is constant.
ρ1
The rate of fluid volume entering one end equals the volume leaving at the other end
Where the pipe narrows (constriction), the fluid moves faster, and vice versa
13
Bernoulli’s Equation – energy conservation
OPTIONAL TOPIC
Daniel Bernoulli, Swiss physicist, 1700 – 1782
•
•
•
An ideal fluid flows through a region where its speed, crosssectional area, and/or altitude changes.
The pressure in the fluid varies with these changes.
Application of energy conservation yields the Bernoulli Equation
ΔW = work done by pressure = ΔEmech = E 2 − E1
Pressure P1
speed v1
height y1
Pressure P2
speed v2
height y2
Equal volumes flow in equal times
due to continuity equation
Volume
ΔV = A1Δx1
Volume
ΔV = A2Δx2
time t
time t + Δt
Bernoulli’s Equation – energy conservation
OPTIONAL TOPIC
The fluid is incompressible. The continuity equation (no leaks) requires:
ΔV = A Δ x = A Δ x
m1 = m 2 ≡ m = ρΔV
1
1
2
2
The net work done on fluid volume by external pressures:
ΔW = F1dx 1 − F2dx 2 = P1A1dx 1 − P2 A 2dx 2 = [P1 − P2 ] ΔV
The change in the mechanical energy:
1
1
ΔE = E2 − E1 = 2 mv 22 − 2 mv12 + mgy 2 − mgy1
Equate to the net work and divide through by the volume ΔV:
P1 − P2 =
Rearrange:
1
2
ρv 2
2
−
P1 +
Pressure P1
speed v1
height y1
1
2
ρv 1
2
1
2
ρv 1
2
+ ρg( y 2 − y1 )
+ ρgy1 = P2 +
1
2
ρv 2
2
Equal volumes flow in equal times
d
due
to continuity
i i equation
i
Volume
ΔV = A1Δx1
+ ρgy 2
Pressure P2
speed v2
height y2
Volume
ΔV = A2Δx2
time t
time t + Δt
14
OPTIONAL TOPIC
Bernoulli’s Equation – energy conservation
P1 +
1
2
ρv 1
2
+ ρgy1 = P2 +
1
2
ρv 2
2
+ ρgy 2
Bernoulli’s Equation is often expressed as:
P+
1
2
ρv
2
+ ρgy = Constant
For a fluid at rest Bernoulli’s Equation reduces to the earlier
result for variation of pressure with depth:
P1 = P2 + ρg[ y 2 - y1 ]
For flow at constant altitude (y1 = y2) the pressure is lower
where ther fluid flows faster:
P1 +
1
2
ρv 1
2
= P2 +
1
2
ρv 2
2
Gases are compressible, but the above pressure / speed
relation holds true.
OPTIONAL TOPIC
Bernoulli effect causes airplane wings to work
•
•
•
•
Streamlines flow around a moving airplane wing
Lift is the upward force on the wing from the air
Drag is the resistance
The lift depends on the speed of the airplane, the area of the wing, its
curvature, and the angle between the wing and the horizontal
Low Pressure where
flow velocity is high
Higher Pressure where
flow velocity is low
15
Lift – General
OPTIONAL TOPIC
• In general, an object moving through a fluid experiences lift
as a result of any effect that causes the fluid to change its
direction as it flows past the object
• Some factors that influence lift are:
–
–
–
–
The shape of the object
The object’s orientation with respect to the fluid flow
Any spinning of the object
The texture of the object’s surface
Golf Ball
•
•
Th ball
The
b ll is
i given
i
a rapid
id backspin
b k i
The dimples increase friction
•
It travels farther than if it was not
spinning
– Increases lift
Atomizer
•
•
•
•
•
OPTIONAL TOPIC
A stream of air passes over one
end of an open tube
The other end is immersed in a
liquid
The moving air reduces the
pressure above the tube
The fluid rises into the air stream
The liquid is dispersed into a fine
spray of droplets
16
17