Fluids mechanics for engineering applications

PHYS222 – LSSU – Bazlur Slide 1
Chapter - 11
Fluids

Fluids
11.1 Mass Density
11.2 Pressure
11.3 Pressure and Depth in a Static Fluid
11.4 Pressure Guages
11.5 Pascal’e Principle
11.6 Archimedes’ Principle
11.7 Fluid in Motion
11.8 The Equation of Continuity
11.9 Bernoulli’s Equation
11.10 Applications of Bernoulli’s Equation
11.11 Viscous Flow

11.1 Mass Density
Definition of Mass Density
The mass density of a substance is the mass of a
substance per unit volume:
V
m


SI Unit of Mass Density: kg / m3

11.1 Mass Density

11.1 Mass Density
Example-1: Blood as a Fraction of Body Weight
The body of a man whose weight is 690 N contains about
5.2x10-3
m3
of blood.
(a) Find the blood’s weight and (b) express it as a
percentage of the body weight.
   kg
5
.
5
m
kg
1060
m
10
2
.
5 3
3
3



 

V
m
V
m



11.1 Mass Density
   N
54
s
m
80
.
9
kg
5
.
5 2


mg
W
(a)
(b) %
8
.
7
%
100
N
690
N
54
Percentage 



11.1 Mass Density
Definition of Weight Density
The Weight density of a substance is the Weight of a
substance per unit volume:
V
mg


SI Unit of Weight Density: N / m3

Specific Gravity
• Is relative density.
• It is a ratio of the densities of two materials.
• Dimensionless.
• Using water as a reference can be convenient, since density of water is
(approximately) 1000 kg/m³ or 1 g/cm³ at 4C.
• Specific gravity of Mercury is 13.6, means it is 13.6 times massive or heavier
than water.
Reference
of
Density
Substance
of
Density
Gravity
Specific 

11.2 Pressure
A
F
P 
SI Unit of Pressure: 1 N/m2
= 1Pa
Pascal

Pressure
• Pressure is force per unit area.
P = F/A
• Pressure is not a vector quantity.
• To calculate the pressure we only consider the
magnitude of the force.
• The force generated by the pressure of the static fluid
is always perpendicular to the surface of the contact.
Pa = N/m2
= 1.450 x 10-4
lb/in2
(very small quantity)
bar = 105
Pa = 100 kPa
Atmospheric pressure is 101.3 kPa
= 1.013 bar = 1.450 lb/in2
= 760 torr
lb/in2
or psi = 6.895 x 103
Pa

Pressure causes Perpendicular Force
The forces of a liquid pressing against a surface add up
to a net force that is perpendicular to the surface.
Components parallel to
the surface cancels out
Components vertical to
the surface adds up

11.2 Pressure
xample-2: The Force on a Swimmer
uppose the pressure acting on the back
a swimmer’s hand is 1.2x105
Pa. The
urface area of the back of the hand is
4x10-3
m2
.
) Determine the magnitude of the force
that acts on it.
) Discuss the direction of the force.

11.2 Pressure
A
F
P 
  
N
10
0
.
1
m
10
4
.
8
m
N
10
2
.
1
3
2
3
2
5





 
PA
F
Since the water pushes perpendicularly
against the back of the hand, the force
is directed downward in the drawing.

Shape of the Dam
• Would you prefer Dam-A or Dam-B?
• Justify your answer.
Dam-A Dam-B

11.2 Pressure
Atmospheric Pressure at Sea Level: 1.013x105
Pa = 1 atmosphere
Weight of the column of the air of 1m2
= 1.013x105
N
Mass of the column of the air of 1m2
= 1.013x105
N/g = 10326.2 kg

0
1
2 



 mg
A
P
A
P
Fy
mg
A
P
A
P 
 1
2

V
m 
Relation between Pressure and Depth
Since the column is in equilibrium,
The sum of the vertical forces
equal to zero

Vg
A
P
A
P 

 1
2
Ah
V 
Ahg
A
P
A
P 

 1
2
hg
P
P 

 1
2
The pressure increment = gh = weight density x height
Pressure at a depth, h = P1 + gh = Atmospheric Pressure + Liquid Pressure
So, Liquid Pressure at a depth, h = gh = weight density x depth

Pressure
Water pressure acts perpendicular to the sides of a
container, and increases with increasing depth.

Conceptual Example-3: The Hoover Dam
Lake Mead is the largest wholly artificial
reservoir in the United States. The water
in the reservoir backs up behind the dam
for a considerable distance (120 miles).
Suppose that all the water in Lake Mead
were removed except a relatively narrow
vertical column.
Would the Hoover Dam still be needed
to contain the water, or could a much less
massive structure do the job?

Example-4: The Swimming Hole
Points A and B are located a distance of 5.50 m beneath the surface
of the water. Find the pressure at each of these two locations.

     
Pa
10
55
.
1
m
50
.
5
s
m
80
.
9
m
kg
10
00
.
1
Pa
10
01
.
1
5
2
3
3
pressure
c
atmospheri
5
2







 

 

P
gh
P
P 

 1
2

Where to put the pump?
• Pump-X pushed the water up.
• Pump-Y removes the air from the
pipe and air pressure outside pushes
the water up.
• Pump-Y has a limit, what is that?
gh
Patm 

Pump-X Pump-Y
 
  
mm
10306
m
306
.
10
s
m
80
.
9
m
kg
10
1.0
Pa
10
01
.
1
2
3
3
5






g
P
h atm

gh
Patm 


Pressure Gauges
• One of the simplest pressure
gauges is the mercury
barometer used for measuring
atmospheric pressure.
• 760 mm of mercury

11.4 Pressure Gauges
gh
Patm 

 
  
mm
760
m
760
.
0
s
m
80
.
9
m
kg
10
13.6
Pa
10
01
.
1
2
3
3
5






g
P
h atm

Point A and B are at the same
depth, so,
PB = PA
Patm = P1 + gh
= 0 + gh

Open-tube manometer
• The phrase “open-tube” refers to
the fact that one side of the U-
tube is open to atmospheric
pressure.
• The tube contains mercury
• Its other side is connected to the
container whose pressure P2 is
to be measured.
PB = PA
P2 = Patm + gh
P2 - Patm = gh
• P2 – Patm is called the guage pressure
• P2 is called the absolute pressure

A
B P
P
P 

2
gh
P
PA 

 1
gh
P
P atm 



 

pressure
gauge
2
absolute pressure

sphygmomanometer
blood-pressure instrument: an instrument used to
measure blood pressure in an artery that consists of a
pressure gauge, an inflatable cuff placed around the
upper arm, and an inflator bulb or pressure pump.
Systolic pressure is 120 mm of mercury
Diastolic pressure is 80 mm of mercury

11.5 Pascal’s Principle
Pascal’s Principle
Any change in the pressure applied
to a completely enclosed fluid is transmitted
undiminished to all parts of the fluid and
enclosing walls.

 
m
0
1
2 g
P
P 


1
1
2
2
A
F
A
F










1
2
1
2
A
A
F
F

Example-7: A Car Lift
The input piston has a radius of 0.0120 m
and the output plunger has a radius of
0.150 m.
The combined weight of the car and the
plunger is 20500 N. Suppose that the input
piston has a negligible weight and the bottom
surfaces of the piston and plunger are at
the same level. What is the required input
force?

   
 
N
131
m
150
.
0
m
0120
.
0
N
20500 2
2
2 



F









1
2
1
2
A
A
F
F

 A
P
P
A
P
A
P
FB 1
2
1
2 



gh
P
P 

 1
2
ghA
FB 

hA
V 
g
V
FB
fluid
displaced
of
mass



Archimedes’ Principle
Any fluid applies a buoyant force to an object that is partially
or completely immersed in it; the magnitude of the buoyant
force equals the weight of the fluid that the object displaces:
 
fluid
displaced
of
Weight
fluid
force
buoyant
of
Magnitude
W
FB 

If the object is floating then the
magnitude of the buoyant force
is equal to the magnitude of its
weight.

Example-9: A Swimming Raft
The raft is made of solid square
pinewood. Determine whether
the raft floats in water and if
so, how much of the raft is beneath
the surface.

   
N
47000
s
m
80
.
9
m
8
.
4
m
kg
1000 2
3
3
max



 g
V
Vg
F water
water
B 

    m
8
.
4
m
30
.
0
m
0
.
4
m
0
.
4 

raft
V

   
N
47000
N
26000
s
m
80
.
9
m
8
.
4
m
kg
550 2
3
3




 g
V
g
m
W raft
pine
raft
raft 
The raft floats!

g
Vwater
water


N
26000
B
raft F
W 
If the raft is floating:
     
2
3
s
m
80
.
9
m
0
.
4
m
0
.
4
m
kg
1000
N
26000 h

     m
17
.
0
s
m
80
.
9
m
0
.
4
m
0
.
4
m
kg
1000
N
26000
2
3


h

Conceptual Example-10: How Much Water is Needed
to Float a Ship?
A ship floating in the ocean is a familiar sight. But is all
that water really necessary? Can an ocean vessel float
in the amount of water than a swimming pool contains?

11.7 Fluids in Motion
In steady flow the velocity of the fluid particles at any point is constant
as time passes.
Unsteady flow exists whenever the velocity of the fluid particles at a
point changes as time passes.
Turbulent flow is an extreme kind of unsteady flow in which the velocity
of the fluid particles at a point change erratically in both magnitude and
direction.

Fluid flow can be compressible or incompressible. Most liquids are
nearly incompressible.
Fluid flow can be viscous or nonviscous.
An incompressible, nonviscous fluid is called an ideal fluid.

When the flow is steady, streamlines are often used to represent
the trajectories of the fluid particles.

Making streamlines with dye
and smoke.

The mass of fluid per second that flows through a tube is called
the mass flow rate.

2
2
2
2
v
A
t
m




1
1
1
1
v
A
t
m





distance
t
v
A
V
m 


 


2
2
2
1
1
1 v
A
v
A 
 
Equation Of Continuity
The mass flow rate has the same value at every position along a
tube that has a single entry and a single exit for fluid flow.
SI Unit of Mass Flow Rate: kg/s

Incompressible fluid: 2
2
1
1 v
A
v
A 
Volume flow rate Q: Av
Q 

Example-12: A Garden Hose
A garden hose has an unobstructed opening
with a cross sectional area of 2.85x10-4
m2
.
It fills a bucket with a volume of 8.00x10-3
m3
in 30 seconds.
Find the speed of the water that leaves the hose
through (a) the unobstructed opening and (b) an obstructed
opening with half as much area.

Av
Q 
   s
m
936
.
0
m
10
2.85
s
30.0
m
10
00
.
8
2
4
-
3
3






A
Q
v
(a)
(b) 2
2
1
1 v
A
v
A 
   s
m
87
.
1
s
m
936
.
0
2
1
2
1
2 

 v
A
A
v

Bernoulli’s Equation
For steady flow,
The speed,
Pressure, and
Elevation
of an incompressible and
nonviscous fluid are related
by Bernoulli’s equation.
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 





The fluid accelerates toward the
lower pressure regions.
According to the pressure-depth
relationship, the pressure is lower
at higher levels, provided the area
of the pipe does not change.

• Bernoulli’s Equation from Work-energy theorem:
• Work done on an object by external non-conservative forces is
equal to the change in total mechanical energy.
• The pressure within a fluid is caused by collisional forces, which
are non-conservative.
• Therefore, when a fluid is accelerated because of a difference
in pressures, work is being done by non-conservative forces.
   
2
2
2
2
1
1
2
1
2
1
2
1
0
f
nc
mgy
mv
mgy
mv
E
E
E
E
W









   
2
2
2
2
1
1
2
1
2
1
nc mgy
mv
mgy
mv
W 



       V
P
P
As
P
s
F
s
F
W 1
2 





 

• Whenever a fluid is flowing in a horizontal pipe and encounters
a region of reduced cross-sectional area, the pressure of the
fluid drops.
• Can be explained by the Newton’s 2nd
law.
• When moving from the wider region 2 to the narrower region 1,
the fluid speeds up according to the equation of continuity.
• According to the 2nd
law, th eaccelerating fluid must be
subjected to an unbalanced force.
• There can be an unbalanced force only if the pressure in region
2 is higher than the pressure in region 1.
   
2
2
2
2
1
1
2
1
2
1
2
1
0
f
nc
mgy
mv
mgy
mv
E
E
E
E
W









• Explanation:
• In figure-b on the top surface of the blue fluid element, the
surrounding fluid exerts a pressure P.
• This pressure give rise to a force, F = P/A
• On the botom of the blue fluid element, the pressure is slightly
higher, P+P.
• As a result, the force on the bottom surface has a magnitude of
F+ F = (P+P)/A.
• The net force pushing the fluid element up the pipe is F=P/A

• When the fluid element moves through its own length s, the
work done is the
W = F s = P A s = P V
The total work done on the fluid element in moving it from region 2
to region 1is the sum of the small increments of work P V
done as the element moves along the pipe.
This sum amounts to Wnc = (P2 – P1)V
Wnc = (P2 – P1)V = E1 – E2

     
2
2
2
2
1
1
2
1
2
1
1
2 mgy
mv
mgy
mv
V
P
P 




     
2
2
2
2
1
1
2
1
2
1
1
2 gy
v
gy
v
P
P 


 




In steady flow of a nonviscous, incompressible fluid, the pressure, the
fluid speed, and the elevation at two points are related by:
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 





The term
has a constant value at all positions in the flow.
constant
2
2
1


 gy
v
P 


reduces to the result for static
fluids (v1=v2), as it is when
the cross-sectional area
remains constant.
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 




gh
P
P
y
y
g
P
P
gy
gy
P
P
gy
P
gy
P

















1
2
2
1
1
2
2
1
1
2
2
2
1
1
)
(

Conceptual Example-14: Tarpaulins and Bernoulli’s Equation
When the truck is stationary, the
tarpaulin lies flat, but it bulges outward
when the truck is speeding down
the highway.
Account for this behavior.

reduces to
When all parts have the same elevation (y1 = y2).
The term P+ ½  remains constant throughout a horizontal pipe.
If  increases, P decreases and vice versa.
2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 




2
2
2
1
2
2
1
2
1
1 v
P
v
P 
 


constant
2
2
1

 v
P 

Example-16: Efflux Speed
The tank is open to the atmosphere at
the top. Find and expression for the speed
of the liquid leaving the pipe at
the bottom.

2
2
2
2
1
2
1
2
1
2
1
1 gy
v
P
gy
v
P 


 




atm
P
P
P 
 2
1
0
2 
v
h
y
y 
 1
2
gh
v 
 
2
1
2
1
gh
v 2
1 

11.11 Viscous Flow
Flow of an ideal fluid.
Flow of a viscous fluid.

11.11 Viscous Flow
Force Needed to Move a Layer of Viscous Fluid with
Constant Velocity
The magnitude of the tangential force required to move a fluid
layer at a constant speed is given by:
y
Av
F


coefficient
of viscosity
SI Unit of Viscosity: Pa·s
Common Unit of Viscosity: poise (P)
1 poise (P) = 0.1 Pa·s

11.11 Viscous Flow
Poiseuille’s Law
The volume flow rate is given by:
 
L
P
P
R
Q


8
1
2
4



11.11 Viscous Flow
Example-17: Giving and Injection
A syringe is filled with a solution whose
viscosity is 1.5x10-3
Pa·s. The internal
radius of the needle is 4.0x10-4
m.
The gauge pressure in the vein is 1900 Pa.
What force must be applied to the plunger,
so that 1.0x10-6
m3
of fluid can be injected
in 3.0 s?

11.11 Viscous Flow
   
 
Pa
1200
m
10
4.0
s
0
.
3
m
10
0
.
1
m
025
.
0
s
Pa
10
5
.
1
8
8
4
4
-
3
6
3
4
2













R
LQ
P
P

11.11 Viscous Flow
Pa
1200
1
2 
 P
P
Pa
1900
1 
P
Pa
3100
2 
P
   N
25
.
0
m
10
0
.
8
Pa
3100 2
5
2 

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