A fluid

baconossifiedMechanics

Oct 29, 2013 (3 years and 5 months ago)

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Chapter 4

fluids


References:

1
-

Physics in Biology and Medicine, 3
rd

E, Paul
Davidovits
.

2
-

College Physics, 6
th

E,
Serway

3
-

Web Sites


30 October 2013

1

Objectives


1
-

Understand the static behavior of fluids



2
-
Illustrate the properties of fluid pressure,
buoyant force in liquids, and surface tension



3
-
Understand the behaviors of fluids in motion



4
-
Study some examples from biology and
zoology.

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2

Introduction


The differences in the physical properties of solids, liquids, and gases
are explained in terms of the forces that bind the molecules.



1


Solids: the molecules are rigidly bound; a solid therefore has a
definite
shape and volume.


2
-
Liquids: The molecules constituting a liquid are not bound together with
sufficient force to maintain a definite shape, but the binding is sufficiently
strong to maintain a
definite volume
.


3
-

Gases: the molecules are not bound to each other. Therefore a gas has
neither a definite shape nor a definite volume



Fluids are liquids and gases


Fluids and solids are governed by the same laws of mechanics.

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3

1
-

Static Fluids

Force and Pressure in a Fluid


When a force is applied to



1
-
A solid:
this force is transmitted to the other
parts of the solid with its direction unchanged.




2
-

A fluid:
Because of a fluid’s ability to flow, it
transmits a force uniformly in all directions.
Therefore, the pressure at any point in a fluid at
rest is the same in all directions.

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4

Pressure


A fluid in a container exerts a force on all parts of the container in
contact with the fluid.


Fluid also exerts a force on any object immersed in it.
P=
F/A


F is always perpendicular to A.




The pressure in a fluid increases

with depth because of the weight

of the fluid above
.




P
2

−P
1

=
ρ
gh

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5

Pressure units

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6


1 torr


=1mm Hg


=13
.5mm water


=1
.33
×
10
3

dyn/cm
2


=1
.32
×
10
−3

atm


=1
.93
×
10
−2

psi


=1
.33
×
10
2

Pa (N/m
2
)



4
-
2
-
b PRESSURE MEASUREMENTS



The open
-
tube manometer


P = P
0

-

rgh.


P

is called the
absolute

pressure
,


P
-

P
0

is called the
gauge pressure
.


If P > P
0

h

is +ve


If P > P
0

h

is
-
ve





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7

Barometer

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8

Pascal’s Principle

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9

In an incompressible liquid, the increase in the pressure at any
point is transmitted undiminished to all other points in the
liquid. This is known as Pascal’s principle.

Archimedes’ Principle



Archimedes’ principle states that


a body
partially

or
wholly

submerged in a fluid
is buoyed upward by a force that is equal in
magnitude to the weight of the displaced fluid.

B=(P
a
-
P
b
)A=(
r
fluid
.g.h)A=
r
fluid

g.V=M
fluid
.g

4
-
2
-
d Archimedes’ Principle


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11

B=(P
a
-
P
b
)A=(
r
fluid
.g.h)A


=
r
fluid

g.V=M
fluid
.g

Archimedes’ principle states that a body
partially or wholly submerged in a fluid is
buoyed upward by a force that is equal in
magnitude to the weight of the displaced fluid.

Case 1
: Totally Submerged Object

B=

r
fluid

g.V
fluid


F
g
=
M
object
.g
=
V
object

r
object
.g


B
-
F
g
=
r
fluid

g.V
fluid
-

V
object

r
object
.g

,
V
fluid
=
V
object
=V



B
-
F
g

=
g.V

(
r
fluid



r
object
)

Case 2

Floating Object


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12


r
fluid

g.V
fluid
=V
object

r
object.
g


the fraction of the volume of a floating object that is below the fluid surface is
equal to the ratio of the density of the object to that of the fluid.

Problem 1



In a huge oil tanker, salt water has flooded an oil tank to a depth of 5.00 m. On
top of the water is a layer of oil 8.00 m deep, as in the cross
-
sectional view of the
tank in Figure. The oil has a density of 0.700 g/cm
3
. Find the pressure at the
bottom of the tank. (Take 1 025 kg/m
3

as the density of salt water.)


Solution


P
1
=P
0
+
r
oil

gh
1
= 1.01x10
5

Pa+(7.00x10
2

kg/m
3
)(9.80m/s
2
)(8.00m)



= 1.56x10
5

Pa


so P
bott
= P1+
r
water

gh
2
= 1.56x10
5

Pa+ (1.025x10
3

kg/m
3
)
)(9.80m/s
2
)(5.00m)



= 2.06x10
5

Pa


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13

Problem 2


Estimate the net force exerted on your eardrum
-

A
~
1 cm
2

-

due to the
water above when you are swimming at the bottom of a pool that is 5.0
m deep.



Solution





DP=P
-
P
0
=
r




=(
1.00x10
3

kg/m
3
)(9.80m/s
2
)(5.00m)= 4.9x10
4
Pa


F
net
= A
D
P=(1x10
-
4

m
2
)( 4.9x10
4
Pa)
~

5N



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14

4
-
3 The human brain


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15

The human brain is immersed in a
fluid (the
cerebrospinal fluid
) of
density
1 007 kg/m
3
, which is
slightly less than the average
density of the brain,
1 040 kg/m
3
.

Most of the weight of the brain is supported
by the buoyant force of the surrounding fluid.

4
-
4 Buoyancy of Fish



The bodies of some fish contain:


either
porous bones


or
air
-
filled swim bladders


that decrease their average density and
allow them to float in water without an
expenditure of energy.


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16

4
-
4 Buoyancy of Fish



EX.
Cuttlefish


contains a porous bone that has a density of 0
.
62 g/cm
3


its body has a density of 1
.
067 g/cm
3
.


the percentage of the body volume occupied by the porous bone that
makes the average density of the fish be the same as the density of sea
water (1
.
026 g/cm
3
) by using the following equation



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-
The cuttlefish lives in the sea at a depth of about 150 m.

At this depth, the pressure is 15 atm .


-
The spaces in the porous bone are filled with gas at a pressure of about
1 atm.


-
Therefore, the porous bone must be able to withstand a pressure of
14 atm.

The cuttlefish alters its density by injecting or
withdrawing fluid from its porous bone.

In fish that possess swim bladders


The decrease in density is provided by the gas
in the bladder.


To achieve the density reduction calculated in
the preceding example, the volume of the
bladder is only about 4% of the total volume
of the fish


Fish with swim bladders alter their density by
changing the amount of gas in the bladder.


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4
-
5 Surface Tension

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19

The surface of a liquid contract and behave somewhat like a
stretched membrane.


This contracting tendency results in a surface
tension that resists an increase in the free surface


That surface tension is force acting tangential to
the surface, normal to a line of unit length on the
surface


F
T

=
TL

capillary action
.


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20

The surface molecules near the wall
are attracted to the wall. This attractive
force is called
adhesion
.


These molecules are also subject to
the
attractive cohesive

force exerted
by the liquid



So

If the adhesive force is greater than the
cohesive force, the liquid wets the container
wall, and the liquid surface near the wall is
curved upward.

If the opposite is the case, the liquid surface is curved downward

F
m

= 2
πRT

capillary action
.

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21

W=

πR
2
h
r
g

F
m

= 2
πRT

F
m

= 2
πRT cos
q
in Y
-
direction

2
πRT
cos
θ
=
πR
2

hρg

h
=2
T
cos
θ / Rρg

r
gh= 2T/R if
q
=0

P
out

P
in

Air

Problem

3


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22


Find the height to which water would rise in a capillary tube with a radius
equal to 5.0 x 10
-
5

m. Assume that the contact angle between the water and the
material of the tube is small enough to be considered zero.


Solution


h=2Tcos
0
0

/
r
gR


=2(0.073 N/m) / (1.00x10
3

kg/m
3
)(9.80 m/s
2
)(5.0x10
-
5

m)


=0.30 m


Assignment


Solve the following problems


1,3,5,6



Surfactants

are molecules that lower surface
tension of liquids. (The word is an abbreviation of
surface active agent
.)



Write a short account on the effect of surfactants
that lowers the surface tension


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