Download Lecture 13: Fluids

Physics 106 Final Exam
 When?
 May. 10 Tuesday, 2:30 — 5:00 pm
 Duration: 2.5 hours
 Where?
 ECEC-100 (SEC-008, 10)
 What?
 Phys-106 (75%), Phys-105 (25%)
 How?




Review sessions (today’s lecture and next week’s recitation)
Equation sheet
Prof. Janow review session on Monday May 9, 3:00-5:00 pm in THL-2
Sample exams on my website: web.njit.edu/~cao/106
 What if?
 28 multiple choice problems (2.5hr/28 ~ 5 min/prob)
 24 correct answers yields a score of 100%
 Today follows Thursday schedule: Chap. 14.1-14.5
 HW13 due by 11:00 pm on May 10
Copyright R. Janow Spring 2010
Physics 106 Lecture 13
Fluid Mechanics
SJ 8th Ed.: Chap 14.1 to 14.5
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What is a fluid?
Pressure
Pressure varies with depth
Pascal’s principle
Methods for measuring pressure
Buoyant forces
Archimedes principle
Fluid dynamics assumptions
An ideal fluid
Continuity Equation
Bernoulli’s Equation
Copyright R. Janow Spring 2010
What is a fluid?
Solids: strong intermolecular forces



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”



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



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

molecules move independently of each other

comparatively easy to compress: density depends on temperature and pressure
Fluid Statics - fluids at rest (mechanical equilibrium)
R. Janow Spring 2010
Fluid Dynamics – fluid flow (continuity, energy Copyright
conservation)
Mass and Density
• Density is mass per unit volume at a point:
m

V
•
•
or
m

V
• scalar
• units are kg/m3, gm/cm3..
• water= 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.
Copyright R. Janow Spring 2010
Force & Pressure
• The pressure P on a “small” area A is the ratio of the
magnitude of the net force to the area
F
P
or P  F / A
A


F  PA  PAn̂
•
•
PA n̂
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:
 1 Pascal (Pa) = 1 Newton/m2 (SI)
 1 PSI (Pound/sq. in) = 6894 Pa.
 1 milli-bar = 100 Pa.
Copyright R. Janow Spring 2010
Forces/Stresses in Fluids
• Fluids do not allow shearing stresses or tensile stresses.
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?
Copyright R. Janow Spring 2010
Pressure in a fluid varies with depth
Fluid is in static equilibrium
The net force on the shaded volume = 0
y=0
Incompressible liquid - constant density 
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Horizontal surface areas = A
Forces on the shaded region:
– Weight of shaded fluid:
–
–
Mg
Downward force on top: F1 =P1A
Upward force on bottom: F2 = P2A
y1
F1
P1
h
F2
y2
P2
Mg
Fy  0  P2A  P1A  Mg
•
In terms of density, the mass of the
shaded fluid is:
The extra pressure at
extra depth h is:
M  V  Ah
P  P2  P1  gh
 P2 A  P1A  ghA
h  y1  y2
Copyright R. Janow Spring 2010
Pressure relative to the surface of a liquid
air
Example: The pressure at depth h is:
P0
Ph  P0  gh
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



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 equations also
hold approximately for
gases such as air if the
density does not vary
much across h
Ph
Copyright R. Janow Spring 2010
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
≡ 1 N/m
bar (bar)
2
10
−5
atmosphere
(atm)
9.8692×10
−6
torr
(Torr)
pound-force per
square inch (psi)
−
7.5006×10
3
−6
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
Copyright R. Janow Spring 2010
Pressure Measurement: Barometer
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near-vacuum
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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  Hggh
Mercury (Hg)
How high is the Mercury column?
P0
1.013  105 Pa
h 

 0.760 m
Hgg (13.6  103 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
 waterg

1.013  105 Pa
(1.0  10 kg / m )(9.80 m/s )
3
3
2
 10.34 m
Height limit for a suction pump
Copyright R. Janow Spring 2010
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.
F
P
Example: open container
•
p0
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.
•
ph
Ph  P0  gh
•
Add piston of area A with lead balls on it & weight W.
Pressure at surface increases by P = W/A
Pext  P0  P
•
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  ghCopyright R. Janow Spring 2010
Pascal’s Law Device - Hydraulic press
A small input force generates a large output force
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•
Assume the working fluid is incompressible
Neglect the (small here) effect of height on pressure
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The volume of liquid pushed down on
the left equals the volume pushed up
on the right, so:
A1x1  A2x2
x 2
A1


x 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, no friction, etc.
Work1  F1x1  Work 2  F2x2
mechanical advantage
F2 x1
A2



F1 Copyright
x 2 R. A
1 Spring 2010
Janow
Archimedes Principle
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C. 287 – 212 BC
Greek mathematician, physicist and engineer
Computed p 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.
Copyright R. Janow Spring 2010
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?
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The extra pressure at the lower surface
compared to the top is:
P  Pbot  Ptop   flgh
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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  PA  flghA  flgV  Mflg
•
The weight of the actual cube is:
Fg  Mcube g  cube gV
The cube rises if B > Fg (fl > cube)
The cube sinks if Fg> B (cube > fl)
Similarly for irregularly shaped
objects
cubeCopyright
is the R.
average
density
Janow Spring
2010
Archimedes's Principle: totally submerged object
Object – any shape - is totally submerged in a fluid of density fluid
The upward buoyant force is the
weight of displaced fluid:
B   fluid gVfluid
The downward gravitational force on the
object is:
Fg  Mobject g  object 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  [ fluid  object ] 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
Copyright R. Janow Spring 2010
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   fluid gVfluid  Fg  object gVobject
Solving:
 fluid Vfluid  object Vobject
object
Vfluid

Vobject
 fluid
 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
Copyright R. Janow Spring 2010
Example: What fraction of an iceberg is underwater?
Apply Archimedes’ Principle
displaced seawater
object
Vunderwater
Vfluid


Vtotal
Vobject
 fluid
glacial fresh water ice
From table:
object  ice  0.917  103 kg/m3
fluid  seawater  1.03  103 kg/m3
Water expands when it freezes. If not....
...ponds, lakes, seas freeze to the bottom in winter

Vunderwater
 89.03%
Vtotal
What if iceberg is in a freshwater lake?
 fluid   freshwater  1.00  103 kg/m3
Vunderwater
 91.7%
Vtotal
Floating objects are more buoyant in saltwater
Freshwater tends to float on top of seawater...
Copyright R. Janow Spring 2010
Fluids’ Flow is affected by their viscosity
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Viscosity measures the internal friction in a fluid.
Low viscosity
• gases
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Medium viscosity
• water
• other fluids that pour
and flow easily
High viscosity
• honey
• oil and grease
• 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:
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•
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The fluid is nonviscous – internal friction is neglected
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
Copyright R. Janow Spring 2010
Flow of an ideal fluid through a short section of pipe
Constant density and velocity within volume element dV
Incompressible fluid means d/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 
 A
length dx
dx
 Av
dt
 dV  Adx
dM d

(V )
dt dt
Imass  mass flow rate  Av
Ivol  volume flow rate  Av
Jmass  mass flow/unit area  v
Copyright R. Janow Spring 2010
Equation of Continuity: conservation of mass
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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  1A1v1
•
 mass flow out  2 A2v2
The fluid is incompressible so:
2
1  2  a cons tan t
 A1v1  A2v2
•
•
This is called the equation of continuity
for an incompressible fluid
The product of the area and the fluid
speed (volume flux) at all points along a
pipe is constant.
1
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
Copyright R. Janow Spring 2010