physics106_lecture13.ppt

Copyright R. Janow Spring 2010
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

Physics 106 Lecture 13
Fluid Mechanics
SJ 8th 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
• Fluid dynamics assumptions
• An ideal fluid
• Continuity Equation
• Bernoulli’s Equation

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
Fluid Statics - fluids at rest (mechanical equilibrium)
Fluid Dynamics – fluid flow (continuity, energy conservation)
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

Mass and Density
• Density is mass per unit volume at a point:
V
m
or
V
m






• 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.

Force & Pressure
• Pressure is a scalar while force is a vector
• The direction of the force producing a pressure is perpendicular
to some area of interest
n̂
A
P
A
P
F 






• The pressure P on a “small” area A is the ratio of the
magnitude of the net force to the area
A
/
F
P
or
A
F
P 



n̂
PA
• At a point in a fluid (in mechanical
equilibrium) the pressure is the same in
any direction
h
Pressure units:
 1 Pascal (Pa) = 1 Newton/m2 (SI)
 1 PSI (Pound/sq. in) = 6894 Pa.
 1 milli-bar = 100 Pa.

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?

Pressure in a fluid varies with depth
Fluid is in static equilibrium
The net force on the shaded volume = 0
• Incompressible liquid - constant density 
• 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
y2 Mg
F1
F2
y=0
P1
P2
h
Mg
A
P
A
P
F 2
y 



 1
0
Ah
V
M 




• In terms of density, the mass of the
shaded fluid is:
The extra pressure at
extra depth h is:
gh
P
P
P 



 1
2
y
y
h 2
1 

ghA
A
P
A
P2 


 1

Pressure relative to the surface of a liquid
gh
P
Ph 

 0
 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
air
Ph
P0
liquid
h
Example: The pressure at depth h is:
 P0 is the local atmospheric (or ambient)
pressure
 Ph is the absolute pressure at depth h
 The difference is called the gauge pressure
P0
Ph
Preceding equations also
hold approximately for
gases such as air if the
density does not vary
much across h

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)
bar (bar) atmosphere
(atm)
torr
(Torr)
pound-force per
square inch (psi)
1 Pa ≡ 1 N/m
2
10
−5
9.8692×10
−6
7.5006×10
−
3
145.04×10
−6
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
• 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.
Mercury (Hg)
near-vacuum
gh
P Hg


0
One 1 atm = 760 mm of Hg
= 29.92 inches of Hg
m
0.760
)
m/s
.
)(
m
/
kg
10
(13.6
Pa
10
1.013
g
P
h 2
3
5
Hg






80
9
3
0
How high is the Mercury column?
m
10.34
)
m/s
.
)(
m
/
kg
10
(1.0
Pa
10
1.013
g
P
h 2
3
5
water






80
9
3
0
How high would a water column be?
Height limit for a suction pump

Pascal’s Principle
• 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.
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.
Example: open container
gh
P
Ph 

 0
F
P
p0
ph
• Add piston of area A with lead balls on it & weight W.
Pressure at surface increases by P = W/A
gh
P
P ext
h 


• Pressure at every other point in the fluid (Pascal’s law),
increases by the same amount, including all locations at
depth h.
P
P
Pext 

 0
Ph

Pascal’s Law Device - Hydraulic press
A small input force generates a large output force
x
A
x
A 2
1 

 2
1
x
F
Work
x
F
Work 2
2
1 



 2
1
1
• Assume no loss of energy in
the fluid, no friction, etc.
• Assume the working fluid is incompressible
• Neglect the (small here) effect of height on pressure
• The volume of liquid pushed down on
the left equals the volume pushed up
on the right, so:
Other hydraulic lever devices
using Pascal’s Law:
• Squeezing a toothpaste tube
• Hydraulic brakes
• Hydraulic jacks
• Forklifts, backhoes
A
A
x
x
1
2
2
1




A
A
x
x
F
F
2
1
1
2
1
2





mechanical advantage

Archimedes Principle 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?
• 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.
ball of liquid
in equilibrium
hollow ball
same
upward force
identical pressures at every point

Archimedes’s principle - submerged cube
• 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:
• The extra pressure at the lower surface
compared to the top is:
fl
top
bot gh
P
P
P 




g
M
gV
ghA
PA
B fl
fl
fl 




 
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 weight of the actual cube is:
gV
g
M
F cube
cube
g 


Similarly for irregularly shaped
objects
cube is the average density
The cube rises if B > Fg (fl > cube)
The cube sinks if Fg> B (cube > fl)

Archimedes's Principle: totally submerged object
Object – any shape - is totally submerged in a fluid of density fluid
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 is less than
the density of the fluid, the unsupported
object accelerates upward
– If the density of the object is more than
the density of the fluid, the unsupported
object sinks
The apparent weight is the
external force needed to
restore equilibrium, i.e.
net
g
apparent F
B
F
W 



The upward buoyant force is the
weight of displaced fluid:
fluid
uid
fl gV
B 

The downward gravitational force on the
object is:
object
object
object
g gV
g
M
F 


The volume of fluid displaced and the
object’s volume are equal for a totally
submerged object. The net force is:
object
object
uid
fl
[
g
net V
g
]
F
B
F 






Archimedes’s Principle: floating object
g
g
net F
B
0
F
B
F 




At equilibrium the upward buoyant
force is balanced by the downward
weight of the object:
An object sinks or rises in the fluid until it reaches equilibrium.
The fluid displaced is a fraction of the object’s volume.
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:
object
object
g
fluid
uid
fl gV
F
gV
B 




Solving:
object
object
fluid
uid
fl V
V 


V
V
uid
fl
object
object
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

Example: What fraction of an iceberg is underwater?
V
V
V
V
uid
fl
object
object
fluid
total
underwater




displaced seawater
glacial fresh water ice
Apply Archimedes’ Principle
3
ice
object kg/m
. 3
10
917
0 




3
seawater
fluid kg/m
. 3
10
03
1 




From table:
%
.
V
V
total
underwater
03
89


What if iceberg is in a freshwater lake?
3
freshwater
fluid kg/m
. 3
10
00
1 




1.7%
V
V
total
underwater
9

Water expands when it freezes. If not....
...ponds, lakes, seas freeze to the bottom in winter
Floating objects are more buoyant in saltwater
Freshwater tends to float on top of seawater...

Fluids’ Flow is affected by their viscosity
• Viscosity measures the internal friction in a fluid.
• 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
Low viscosity
• gases
Medium viscosity
• water
• other fluids that pour
and flow easily
High viscosity
• honey
• oil and grease
• glass
Ideal Fluids – four approximations to simplify the
analysis of fluid flow:
• 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

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”
length dx
velocity v
cross-section area A
Av
dt
dx
A
)
V
(
dt
d
dt
dM
I rate
flow
mass
mass








Adx
dV
cylinder
in
fluid
f
o
volume 

Av
I rate
flow
mass
mass 


Av
I rate
flow
volume
vol 

v
J area
flow/unit
mass
mass 



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.
1
2
2
2
2
1
1
1 v
A
v
A out
flow
mass
in
flow
mass 




• The fluid is incompressible so:
t
tan
cons
a



 2
1
v
A
v
A 2
2
1
1 

• 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.
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

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