Transcript Lecture 9: Solids and Fluids

Chapter 9: Solids and Fluids
States of Matter
 Three
states of matter
• Normally matter is classified into one of three (four) states:
solid, liquid, gas (, plasma).
solid : crystalline solid (salt etc.)
amorphous solid (glass etc.)
ordered structure
atoms arranged at almost at random
States of Matter
 Three
(four) states of matter (cont’d)
• Normally matter is classified into one of three (four) states:
solid, liquid, gas (, plasma).
liquid : A molecule in a liquid does random-walk through a series
of interactions with other molecules.
- For any given substance, the liquid
state exists at a higher temperature
than the solid state.
-The inter-molecular forces in a liquid
are not strong enough to hold molecules together in fixed position.
-The molecules wander around in
random fashion.
States of Matter
 Three
(four) states of matter (cont’d)
• Normally matter is classified into one of three (four) states:
solid, liquid, gas (, plasma).
gas : In gaseous state, molecules are in constant random motion
and exert only weak forces on each other.
-The average distance between the molecules of a gas is quite
large compared with the size of molecules.
- Occasionally the molecules collide with each other, but most of
them move freely.
- Unlike solids and liquids, gases can be easily compressed.
plasma : At high temperature, electrons of atoms are free from
nucleus. Such a collection of ionized atoms with equal amounts of
positive (nucleus) and negative charges (electrons) forms a state
called plasma.
Deformation of Solids
 Stress,
strain and elastic modulus
• Until external force becomes strong enough to deform permanently
or break a solid object, the effect of deformation by the external
force goes back to zero when the force is removed – Elastic behavior.
• Stress
: the force per unit area causing a deformation
Strain
: a measure of the amount of the deformation
Elastic modulus : proportionality constant, similar to a spring constant
stress = elastic modulus x strain
Deformation of Solids
 Young’s
modulus: elasticity in length
• Consider a long bar of cross-sectional area A and length L0,
clamped at one end. When an external force F is applied along
the bar, perpendicular to the cross section, internal forces in the
bar resist the distortion that F tends to produce.
• Eventually the bar attains an
equilibrium in which:
(1) its length is greater than L0
(2) the external force is balanced
by internal forces.
The bar is said to be stressed.
F
L
Y
A
L0
tensile stress Young’s modulus
SI unit: Pa = 1 N/m2
tensile strain SI unit: dimensionless
Deformation of Solids
 Young’s
modulus: elasticity in length (cont’d)
• Typical values
• Stress vs. strain
Deformation of Solids
 Shear
modulus: Elasticity of shape
• Another type of deformation occurs when an object is subjected to
a force F parallel to one of its faces while the opposite face is held
fixed by a second force.
• The stress in this
situation is called
a shear stress.
F
x
S
A
h
shear stress Shear modulus
SI unit: Pa = 1 N/m2
shear strain
SI unit: dimensionless
Deformation of Solids
 Bulk
modulus: Volume elasticity
• Suppose that the external forces acting on an object are all
perpendicular to the surface on which the force acts and are
distributed uniformly.
• This situation occurs when a
object is immersed in a fluid.
V
P   B
V
volume stress bulk modulus
SI unit: Pa = 1 N/m2
volume strain SI unit: dimensionless
Deformation of Solids
 An
example
• Example 9.3 : Stressing a lead ball
A solid lead sphere of volume 0.50 m3, dropped in the ocean, sinks
to a depth of 2.0x103 m, where the pressure increases by 2.0x107 Pa.
Lead has a bulk modulus of 4.2x1010 Pa. What is the change in
volume of the sphere?
P
B
V / V
VP
(0.50 m3 )( 2.0 107 Pa)
4
3
V  



2
.
4

10
m
B
4.2 1010 Pa
Density and Pressure
 Density
• The density r of an object is defined as:
r
M
V
M: mass, V: volume
• The specific gravity of a
substance is the ratio of
its density to the density
of water at 4oC, which is
1.0x103 kg/m3, and it is
dimensionless.
SI unit: kg/m3 (cgs unit: g/cm3 )
Density and Pressure
 Pressure
• Fluids do not sustain shearing stresses, so
the only stress that a fluid can exert on a
submerged object is one that tends to
compress it, which is a bulk stress.
• The force F exerted by the fluid on the
object is always perpendicular to the
surfaces of the object.
• If F is the magnitude of a force exerted
perpendicular to a given surface of area A,
then the pressure P is defined as:
P
F
A
F: force, A: area
SI unit: Pa = N/m2
Density and Pressure
 Variation
of pressure with depth
• When a fluid is at rest in a container, all portions of the fluid must
be in static equilibrium – at rest with respect to the observer.
• All points at the same depth must be at the same pressure. If this
were not the case, fluid would flow from the higher pressure region
to the lower pressure region.
• Consider an object at rest
with area A and height h in
a fluid.
P2 A  P1 A  Mg  0
M  rV  rA( y1  y2 )
P2  P1  rg  y1  y2 
• Effect of atmospheric pressure:
P  P0  rgh
P1  P2 for y1  y2  0
P0 : atmospheric pressure, P: pressure at depth h
Density and Pressure
 Examples
• Example 9.5 : Oil and water
P1  P0  rgh1  1.01105 Pa
 (7.00 10 2 kg/m 3 )(9.80 m/s 2 )
(8.00 m)
 1.56 10 Pa
5
Pbot  P1  rgh2
 2.06 105 Pa
r=0.700 g/cm3
h1=8.00 m
r=1025 kg/m3
h2=5.00 m
Density and Pressure
 Pascal’s
principle
• A change in pressure applied to an enclosed fluid is transmitted
undiminished to every point of the fluid and to the walls of the
container.
 Hydraulic
press
P1  P2
F1 F2

A1 A2
F2  F1
A2
A1
F2 > F1 if A2 > A1
Density and Pressure
 Car
lift
• Example 9.7 : Car lift
(a) Find necessary
force by compressed air at piston 1.
2
A1 r1
F1  F2  2 F2
A2 r2
weight=13,300 N
 1.48 103 N
(b) Find air pressure.
F1
P
 1.88 105 Pa
A1
circular x-sec
(c) Show the work done
by pistons is the same.
V1  V2  A1x1  A2x2
A2 / A1  x1 / x2
F1 / F2  A1 / A2
r1=5.00 cm
r2=15.0 cm
W1 F1x1  A1  A2 

     1
W2 F2 x2  A2  A1 
Pressure Measurements
 Absolute
and gauge pressure
• An open tube manometer (Fig.(a))
measures the gauge pressure P-P0
P : absolute pressure
P  P0  rgh
P  P0  rgh
P=PA=PB
P0 : atmospheric pressure
• A mercury barometer (Fig.(b))
measures the atmospheric pressure
P0  P  P0  rgh
• One atmospheric pressure
defined as the pressure equivalent of a
column of mercury that is exactly 0.76 m
in height.
P0  rgh  1.013 105 Pa  1 atm
vacuum
Pressure Measurements
 Blood
pressure measurement
• A specialized manometer
(sphygmomanometer)
-A rubber bulb forces air into a cuff wrap.
-A manometer is attached under cuff and
is under pressure.
-The pressure in the cuff is increased until
the flow of blood through brachial artery
is stopped.
-Then a valve on the bulb is opened, and
measurer listens with a stethoscope to
the artery at a point just below the cuff.
-When the pressure at the cuff and the
artery is just below the max. value
produced by heart (the systolic pressure),
the artery opens momentarily on each beat.
-At this point, the velocity of the blood is high,
and the flow is noisy and can be heard…
Buoyant Forces and
Archimedes’s Principle
 Archimedes’s
principle
Any object completely or partially submerged in a fluid is buoyed
up by a force with magnitude equal to the weight of the fluid
displaced by the object.
Upward force (buoyant force) :
( P2  P1 ) A  r fluidhAg
 r fluidVg
 r fluidV fluid g
B
Downward force:
M obj g  r obj ghA  r objVg
Buoyant Forces and
Archimedes’s Principle
 Archimedes’s
principle and a floating object
Upward force (buoyant force) :
B  r fluidV fluid g
Vobj
Downward force:
M obj g  r objVobj g
r obj V fluid
r fluidV fluid g  r objVobj g 

r fluid Vobj
Vfluid
Buoyant Forces and
Archimedes’s Principle
 Examples
• Example 9.8 : A fake or pure gold crown?
Is the crown made of pure
gold?
Tair =7.84 N
T  mg  0
air
Twater =6.86 N
Twater  mg  B  0
Twater  Tair  B  0
B  Tair  Twater  r water gVwater
 0.980 N
4
Vwater  Vcrown  1.00 10 m
m  Tair / g  0.800 kg
3
rgold=19.3x103 kg/m3
r crown  m / Vcrown  8.00 103 kg/m 3
Buoyant Forces and
Archimedes’s Principle
 Examples
• Example 9.9 : Floating down the river
What depth h is the bottom of
the raft submerged?
rwood=6.00x102 kg/m3
B  mraft g  0  B  mraft g
mraft g  ( r raftVraft ) g
B  mwater g  ( r waterVwater ) g  ( r water Ah) g
( r water Ah) g  ( r waterVraft ) g
r raftVraft
h
 0.0632 m
r raft A
A=5.70 m2
Fluid in Motion
 Some
terminology
• When a fluid is in motion:
(1) if every particle that passes a particular point moves along exactly
the same smooth path followed by previous particles passing the
point, this path is called streamline. If this happens, this flow is said
to be streamline or laminar.
(2) the flow of a fluid becomes irregular, or turbulent, above a certain
velocity or under any conditions that can cause abrupt changes in
velocity.
• Ideal fluid :
1. The fluid is non-viscous :
There is no internal friction force between adjacent layers.
2. The fluid is incompressible :
Its density is constant.
2. The fluid motion is steady :
The velocity, density, and pressure at each point in the fluid do not
change with time.
3. The fluid moves without turbulence :
Each element of the fluid has zero angular velocity about its center.
Fluid in Motion
 Equation
of continuity
• Consider a fluid flowing through a pipe of non-uniform size. The
particles in the fluid move along the streamlines in steady-state flow.
In a small time interval t, the
fluid entering the bottom end of
the pipe moves a distance:
x1  v1t
The mass contained in the bottom
blue region :
M1  r1 A1x1  r1 A1v1t
From a similar argument :
M 2  r2 A2x2  r2 A2v2t
Since M1=M2 (flow is steady):
M1  M 2  r1 A1v1  r2 A2v2
A1v1  A2v2 if incompress ible
Equation of continuity
Fluid in Motion
 An
example
• Example 9.12 : Water garden
Fluid in Motion
 Bernoulli’s
equation
• Consider an ideal fluid flowing through a pipe of non-uniform size.
Work done to the fluid at Point 1
during the time interval t:
W1  F1x1  P1 A1x1  P1V
Work done to the fluid at Point 2
during the time interval t:
W2  P2 A2x2  P2V
Work done to the fluid :
W fluid  P1V  P2V
Fluid in Motion
 Bernoulli’s
equation (cont’d)
If m is the mass of the fluid passing
through the pipe in t , the change
in kinetic energy is:
1 2 1 2
KE  mv2  mv1
2
2
The change in gravitational
potential energy in t is:
PE  mgy2  mgy1
From conservation of energy:
W fluid  KE  PE
Fluid in Motion
 Bernoulli’s
equation (cont’d)
From conservation of energy:
W fluid  KE  PE
( P1  P2 )V 
1 2 1 2
mv2  mv1  mgy2  mgy1
2
2
r  m /V
1 2
1 2
P1  rv1  rgy1  P2  rv2  rgy2  const .
2
2
1 2
P  rv  rgy  const .
Bernoulli’s equation
2
Fluid in Motion
 Venturi
tube
Consider a water flow through
a horizontal constricted pipe.
A1v1  A2v2
A1  A2  v2  v1
y1  y2
1 2
1 2
P1  rv1  P2  rv2
2
2
Fluid in Motion
 Examples
• Example 9.13 : A water tank
Consider a water tank with a hole.
(a) Find the speed of the water
leaving through the hole.
P0 
h =0.500 m
y1 =3.00 m
1 2
rv1  rgy1  P0  rgy2
2
v1  2 g ( y2  y1 )  2 gh
 3.13 m/s
(b) Find where the stream hits the ground.
1 2
y  0  y1   gt  v0 y t  t  0.782 s
2
x  v0 xt  v1t  2.45 m
y
x
Fluid in Motion
 Examples
• Example 9.14 : Fluid flow in a pipe
Find the speed at Point 1.
A1v1  A2v2
v2 
A2=1.00 m2
A1=0.500 m2
h =5.00 m
A1
v1
A2
1 2
1 2
P0  rv1  rgy1  P0  rv2  rgy2
2
2
2
P0 
1 2
1 A 
rv1  rgy1  P0  r  1 v1   rgy2
2
2  A2 
  A 2 
2 gh
v12 1   1    2 g ( y2  y1 )  2 gh  v1 
 11.4 m/s
  A2  
1  ( A1 / A2 ) 2
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Surface
tension
• The net force on a molecule at A is zero
because such a molecule is completely
surrounded by other molecules.
• The net force on a molecule at B is downward
because it is not completely surrounded by
other molecules. There are no molecules
above it to exert upward force. this asymmetry
makes the surface of the liquid contract and
the surface area as small as possible.
• The surface tension is defined as :
F where the surface tension force F

L is divided by the length L along
which the force acts.
SI unit : N/m=(N m)/m2=J/m2
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Surface
tension (cont’d)
• The surface tension of liquids
decreases with increasing
temperature, because the faster
moving molecules of a hot liquid
are not bound together as strongly
as are those in a cooler liquid.
• Some ingredient called surfactants
such as detergents and soaps decrease surface tension.
• The surface tissue of the air sacs in the lungs contain a fluid that has
a surface tension of about 0.050 N/m. As the lungs expand during
inhalation, the body secretes into the tissue a substance to reduce
the surface tension and it drops down to 0.005 N/m.
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Surface
of liquid
• Forces between like-molecules such as between water molecules are
called cohesive forces.
• Forces between unlike-molecules such as those exerted by glass on
water are called adhesive forces.
• Difference in strength between cohesive and adhesive forces creates
the shape of a liquid at boundary with other materials.
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Viscous
fluid flow
• Viscosity refers to the internal friction of a fluid. It is very difficult for
layers of a viscous fluid to slide past one another.
• When an ideal non-viscous fluid flows
through a pipe, the fluid layers slide
past one another with no resistance.
• If the pipe has uniform cross-section
each layer has the same velocity.
• The layers of a viscous fluid have
different velocities. The fluid has the
greatest velocity at the center of the
pipe, whereas the layer next to the wall
does not move because of adhesive
forces between them.
ideal fluid, non-viscous
viscous fluid
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Viscous
fluid flow
• Consider a layer of liquid between two solid surfaces. The lower surface
is fixed in position, and the top surface moves to the right with a velocity
v under the action of an external force F.
• A portion of the liquid is distorted from
its original shape, ABCD, at one instance
to the shape AEFD a moment later. The
force required F to move the upper plate
at a fixed speed v is :
F h
Av
d
where h is the coefficient of viscosity of
the fluid, and A is the area in contact with fluid.
h
SI unit : N s/m2
cgs unit: dyne s/cm2= poise
1 poise=10-1 N s/m2
1 cp (centipoise) = 10-2 poise
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Poiseuille’s
law
• Consider a section of tube of
length L and radius R
containing a fluid under
pressure P1 at the left end and
a pressure P2 at the right.
• Poiseuille’s law describes the
flow rate of a viscous fluid
under pressure difference:
V R 4 ( P1  P2 )

t
8hL
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Reynolds
number
• At sufficiently high velocities, fluid flow changes from simple streamline
flow to turbulent flow, characterized by a highly irregular motion of the
fluid. Experimentally the onset of the turbulence in a tube is determined
by a dimensionless factor called Reynolds number, RN, given by:
rvd
RN 
h
r : density of fluid
v : average speed of the fluid along the direction of flow
d : diameter of tube
h : viscosity of fluid
• If RN is below about 2000, the flow of fluid through a tube is streamline.
• If RN is above about 3000, the flow of fluid through a tube is turbulent.
• If RN is between 2000 and 3000, the flow is unstable.
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Examples
• Example 9.18 : A blood transfusion
A patient receives a blood transfusion through a needle of radius 0.20
mm and length 2.0 cm. The density of blood is 1,050 kg/m3. The bottle
supplying the blood is 0.50 m above the patient’s arm. What is the rate
of the flow through the needle?
P1  P2  rgh  (1050 kg/m 3 )(9.80 m/s 2 )(0.50 m)  5.15 103 Pa
V R 4 ( P1  P2 )

 6.0 108 m3 / s
t
8hL
Surface Tension, Capillary
Action, and Viscous Fluid Flow
 Examples
• Example 9.19: Turbulent flow of blood
Determine the speed at which blood flowing through an artery of
diameter 0.20 cm will become turbulent.
h ( RN )
(2.7 103 N s/m 3 )(3.00 103 )
v

 3.9 m/s
3
3
2
rd
(1.05 10 kg/m )(0.20 10 m)
Transport Phenomena
• A fluid can move place to place as a result of difference in
concentration between two points in the fluid. There are two
processes in this category : diffusion and osmosis.
 Diffusion
• In a diffusion process, molecules move from a region where their
concentration is high to a region where their concentration is lower.
• Consider a container in which a
high concentration of molecules
has been introduced into the left
side (the dashed line is an
imaginary barrier).
All the molecules move in random
direction. Since there are more
molecules on the left side, more
molecules migrate into the right
side than otherwise. Once a
concentration equilibrium is reached, there will be no net movement.
Transport Phenomena
 Diffusion
(cont’d)
• Fick’s law
mass M
 C2  C1 
Diffusion rate 

 DA

time
t
 L 
where D is a constant of proportion called
the diffusion coefficient (unit : m2/s), A is
the cross-sectional area, (…) is the change
in concentration per unit distance
(concentration gradient), and M/t is the
mass transported per unit time. The
concentrations, C1 and C2 are measured
in unit of kg/m3.
Transport Phenomena
 Size
of cells and osmosis
• Diffusion through cell membranes is vital in supplying oxygen to the
cells of the body and in removing carbon dioxide and other waste
products from them.
• A fresh supply of oxygen diffuses from the blood, where its
concentration is high, into the cell, where its concentration is low.
• Likewise, carbon dioxide diffuses from the cell into the blood where
its concentration is lower.
• A membrane that allows passage of some molecules but not others
is called a selectively permeable membrane.
• Osmosis is the diffusion of water across a selectively permeable
membrane from a high water concentration to a low water
concentration.
Transport Phenomena
 Motion
through a viscous medium
• The magnitude of the resistive force on a very small spherical
object of radius r moving slowly through a fluid of viscosity h with
speed v is given by:
resistive
Stokes’s law
frictional
Fr  6hrv
force
• Consider a small sphere of radius r falls
through a viscous medium.
4 3
w  rgV  rg  r 
3

4

B  r f gV  r f g  r 3 
3

force of
buoyant gravity
force
Transport Phenomena
 Motion
through a viscous medium (cont’d)
• At the instance the sphere begins to fall, the force of friction is
zero because the velocity of the sphere is zero.
• As the sphere accelerates, its speed increases resistive
and so does Fr.
frictional
• When the net force goes to zero, the speed
force
of the sphere reaches the so-called terminal
speed vt.
Fr  B  w
4 3
4 3
6hrvt  r f g  r   rg  r 
3

3

2r 2 g
vt 
(r  r f )
9h
force of
buoyant gravity
force
Transport Phenomena
 Sedimentation
and centrifugation
• If an object is not spherical, the previous argument can still be
applied except for the use of Stokes’s law. In this case, we assume
that the relation Fr=kv holds where k is a coefficient.
Fr  B  w Terminal speed condition
rf
B  r f gV 
mg (V  m / r )
r
Fr  kvt
rf
mg  r f
1 
mg 
mg  kvt  vt 
r
k 
r



Transport Phenomena
 Sedimentation
and centrifugation (cont’d)
• The terminal speed for particles in biological samples is usually
quite small; the terminal speed for blood cells falling through plasma
is about 5 cm/h in the gravitational field of Earth.
• The speed at which materials fall through a fluid is called
sedimentation rate. The sedimentation rate in a fluid can be
increased by increasing the effective acceleration g: for example
by using radial acceleration due to rotation
(centrifuge).
v2
ac 
  2r
r
mac  r f
1 
vt 
k 
r

m 2 r  r f
  vt 
1 
k 
r



