Download CHAPTER 7 FLUIDS

CHAPTER 7
FLUIDS
7.1 Density and Pressure

The density of an object of total mass M and volume V is defined as mass
per unit volume and is given by:


mass
M

volume V
SI units : kg/m 3 (g/cm3);
(1 g/cm3=1000 kg/m3)
eg: ρwater = 1.00 x 103 kg/m

3
The pressure on the object is defined to be the force per unit area:
P = F /A

SI Unit : N/m2 or Pascals ( Pa ) where:
1 Pa = 1 N/m
2
 Other units :
-atmosphere :1 atmosphere = 1.01 x 105 Pa
-mm of mercury(Torr):
1 Torr=133.3 Pa
-pounds per sq. inch:
1 lb/in2=6.9x103 Pa
Example: A car weighing 1.2 x 104 N rests on four tires. If the gauge pressure in
each tire is 200 kPa, what is the area of each tire in contact with the road?
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Solution:
Each tire must carry a weight of:
F = 1.2 x 104 / 4 = 0.3 x 104 N.
The
definition
of
pressure
is
Force
4
3
2
P=F/A A=F/P=0.3x10 /200x10 =0.015 m
per
Area.
Therefore:
7.2 Variation of Pressure with Depth
Pressure increases with depth

Eg : diving
With increasing altitude, your ears may “pop” because of reduced external

air pressure.
Eg : fly in the plane / ride in a car going up a mountain.

How the pressure in a fluid varies with depth?
F  w  mg  Vg  ghA  pA
2
The pressure at a depth h due to the weight of the column
p  gh
In an open surface, total pressure / absolute pressure on an object submerged at a
depth h below the surface of a fluid is given by pressure-depth equation.
p  p0  gh
p0 = pressure applied to the fluid surface ( at h=0 )
  = density of the fluid.
On the surface of the Earth: p0=pa = 1.01 x 105 Pa,
where pa=atmospheric pressure (barometer)
Example: A Scuba Diver: Pressure and Force
(a)
What is the total pressure on the back of a scuba diver in a lake at a
depth of 8.00 m?
(b)
What is the force on the diver‟s back due to water alone, taking the
surface of the back to be a rectangle 60.0 cm by 50.0 cm?
Solution
(a)
pt  pa  pw  pa  gh
= 1.01 x 105 +( 1 x 103 ) (9.8) (8)
(b)
F  pw A  ghA
= ( 1 x 103 )(9.8)(8)(60 cm x 50 cm)
PASCAL’S PRINCIPLE
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 Any external pressure applied to an enclosed fluid is transmitted undimished to
every point in the fluid and to the walls of its container.
Application : Hydraulic press and lift (used to raise automobiles and other heavy
objects)
Using the Pascal‟s Principle


pi  po
Fi F0

Ai A0
 A0 
 F0    Fi
 Ai 
Hydraulic force multiplication
Since A0 > Ai , then F0 > Fi (The input force is greatly multiplied)
Advantages of Pascal‟s Principle :
1.
Transmit force from one place to another place.
2.
Multiply that force.
Example: In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the
large piston has a radius of 15 cm. What force must be applied on the small piston
in order to lift a car weighing 20 000 N on the large piston?
Solution :
4
According to Pascal‟s Principle, the external pressure exerted at the small piston is
transmitted
undiminished
to
the
large
piston.
Therefore
F1 F2
p1  p2 

A1 A2
So
2
 A1 
 r12 
r 
 0.050 
F1    F2   2  F2   1  F2  
 (20000 N )  2200 N
A

r
r
0
.
15


 2
 2 
 2
7.5
2
Pressure measurement
It is possible to measure pressure by relating pressure differences to the
variation in the height of a column of liquid. The devices for measuring pressure
are:
1. Open-tube manometer
Figure 9.5: Open tube manometer
5
The pressure P of the gas in the closed cylinder on the left is related to
atmospheric pressure P0
P - P0 =
where

gh
is the density of the fluid in the closed tube.
The pressure P of the gas is called the Absolute Pressure, to distinguish it from
what is directly measured from the height of the column, namely the Gauge
Pressure.
Thus:
P absolute = P gauge + P0.
2.
Tire gauge – to measure air pressure in automobile tires
3.
Barometer – to measure atmospheric pressure
Mercury Barometer
Figure 9.6: Mercury barometer
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

Atmospheric pressure is measured by ensuring that the pressure at the
closed end of the barometer tube is zero (i.e. it is vacuum). Thus
P0 = ρ gh
where ρ is the density of the mercury.
Example: The same car as in previous example sits on a hydraulic press. If the area
of the cylinder holding the car up is 4 times greater than the area of the cylinder
on the other side of the press, what is the force that must be applied to the other
side of the hydraulic press.
Solution:
P = F1/ A1 = F2 / A2
F1 = F2 ( ¼ ) = (1.2 x104 (9.81) ) / 4
= 2.9 x 104 N
7.6 Buoyant Forces
Why things float???
 An object that is either partially or completely submerged in a fluid will
experience an upward buoyant force exerted by the fluid.
 Archimedes’ Principle states that the buoyancy force is equal to the
weight of the fluid displaced.
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Fb  w f  m f g   f gV f
Buoyant force may depend on shape.
Example:
(a)
Calculate the buoyant force of 1.00 x 107 kg of steel completely
submerged in water, and compare this with the steel‟s weight.
(b)
What is the maximum buoyant force that water could exert on this same
steel if it were shaped into a boat that could displace 1.00 x 10 5 m3 of
water.
(Given
 steel  7.8x103 kg / m3 )
Solutions:
(a)
Given msteel =1.00 x 107 kg
FB  mw g   wVw g  1.282 x106 x9.8  1.282 x107 N
8
8
F

m
g


V
g

1
.
00
x
10
x
9
.
8

9
.
8
x
10
N
(b) B
w
w w
BUOYANCY AND DENSITY
Archimedes’ Principle is very useful in determining the volume of an object and
therefore its density.
Eg: crown of king – made of pure or not made of pure gold.
Example: A queen‟s gold crown has mass 1.30kg. However when it is weighed while it
is completely immersed in water, its effective mass is 1.14 kg, is the crown
solid?[  gold
 19.3x103 kg / m3 ]
Solution: From Archimedes‟ Principle
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Weff  Wair  FB
FB=1.30(9.8)-(1.14)(9.8)=1.57N
FB  mw g  wVg
V 
1.57
4 3

1
.
6
x
10
m
3
10 x9.8
m
crown   8.12 x103 kg / m3
V
 It is not solid gold.
How to relate density and floating?
If
object <  fluid
 Object will float
2. If
object >  fluid
 Object will sink
3. If
object =  fluid
1.
 Object will be in equilibrium at any
submerged depth in a fluid.
Example: An aluminum object has a mass of 27.0 kg and a density of 2.70 x 10 3
kg/m 3. The object is attached to a string and immersed in a tank of water.
Determine the
a) volume of the object
b) the tension in the string when it is completely immersed.
Solution:
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a)
The
 Al 
b)
density
mAl
VAl
 V Al 
of
27
2.7 x10
aluminum
is
 0.01m 3
3
Applying Newton's second law for the aluminum block and noting that it is
completely
submerged
and
in
equilibrium,
we
have:
 F  T  B  mg  0
T  mg  B
T  mg   water gVwater
T  (27)(9.8)  (1x10 3 )(9.8)(0.01)
T  166.6 N
7.7 Fluids in Motion (Fluid Dynamics)
There are four characteristics of an ideal fluid flow:
1.
The fluid is nonviscous;
 no external friction force between adjacent fluid layers.
2.
The fluid is incompressible
 the fluid density is constant.
3.
The fluid motion is steady
 velocity, density, and pressure at each point in the fluid do not change in
time.
4.
The fluid moves without turbulence
 each element of the fluid has zero angular velocity about its center – no
eddy currents present in the moving fluid.
-Ideal fluid flow is not characteristic of most real situations.
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Fluids which have the „ideal‟ properties obey 2 important equations:
1.
Equation of continuity
2.
Bernoulli‟s equation
Definition: Flow Rate
The flow rate at each point along a (non-uniform) pipe or hose is defined as the
volume of fluid passing through the pipe at that point per unit time:
F
V
x
A
 A
t
t
where : A is the cross-sectional area of the pipe at that point
v = x/ t is the fluid velocity.
3
SI unit : m /s
Example: How many cubic meters of blood does the heart pump in 75 years life
time, assuming the average flow rate is 5.00L/min.
(1 L = 10-3 m3)
Solution :
From equation
F
V
x
A
 A
t
t
3
3
L
min
3 m
5 m
F  5.0
x10
x1
 8.33x10
min
1L
60s
s
V  8.33x10 5 x75 yx365
d
h
min
s
x 24
x60
x60
1y
1d
1h
1min
= 1.97 x 105 m3
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Idea: If the fluid is incompressible, the flow rate must be the same everywhere
along the pipe (Conservation of fluid particles: what goes in must come out). This
leads directly to the
Continuity Equation:
Equation of continuity : the flow rate through a pipe is constant i.e. the product of
the cross-sectional area of the pipe and the speed of the fluid is constant.
A1
A2
v1
v2
- If the Fluid is incompressible, then the same amount of fluid flows past through
any point in the tube in a given time.
F 1 =F 2
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A1v1=A2v2

A 
2   1 1
 A2 
Av = constant
Example: A nozzle with a radius of 0.250 cm is attached to a garden hose with a
radius of 0.90cm. The flow rate through hose and nozzle is 0.5 liters per second.
Calculate the velocity of water:
a) in the hose
b) in the nozzle
Solution:
hose
Nozz1e
L 103 m3
F  0.5 x
s
1L
= 5 x 10-4 m3/s
F  A1v1
5 x104
v1 
r 2
= 1.96 m/s velocity in the hose
From equation of continuity
A1v1=A2v2
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v2 
A1
xv1
A2
= 25.46 m/s velocity in nozzle
Thus, from this eg. it will produces faster stream if we constrict the flow to
narrower tube.
BERNOULLI’S EQUATION
– the sum of the pressure P, KE per unit volume (  2 / 2 ), and the PE per unit
volume ( gh ) has a constant value at all points along a streamline.
When a fluid flows into narrower channel, its velocity increases. Bernoulli‟s
equation is the direct result of the application of conservation energy to fluid. It is
written as
p +½ ρv2 + ρgh = constant
p1 +½ ρv1 2 + ρgh1= p2 +½ ρv22 + ρgh2
Ho
w??
If the flow is at a constant height, h = constant,
Then
p + ½ ρv2 = constant
Note: If there is no height difference, Bernouilli's equation relates the pressure
at two points along the flow to the fluid velocities:
P1 - P2 = ½ ρ(v22 - v12)
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Example 15:
a) An ideal fluid flows at 4.0 m/s in a horizontal circular pipe. If the pipe narrows
to half of its original radius, what is the flow speed in narrow section?
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b) If the fluid is water and the pressure at the narrow section is 1.8 x 10 5Pa, what
is the pressure at the wide section?
Solution:
A1v1=A2v2
a)
Av
r1 v1
v2  1 1 
 16 m s
A2
r2 2
2
b) Since the fluid is flowing horizontally, h is constant, so Bernoulli‟s Equation can
be written as
p1 +½ ρv12 = p2 +½ ρv2
2
So, p1 = p2 + ½ (v22-v12)
= 1.8 x105 + ½(1000)(162-42)
= 3.0 x 105 Pa.
7.8 Other applications of Bernoulli’s equation
i)
The circulation of air around a thrown baseball
ii)
Atomizers in perfume bottles and paint sprayers
iii)
Vascular flutter and aneurysms
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