Basics of Fluids

MFMcGraw-PHY 1401 Ch09e - Fluids-Revised: 7/12/2010 2
Chapter 9: Fluids
• Introduction to Fluids
• Pressure
• Measurement of Pressure
• Pascal’s Principle
• Gravity and Fluid Pressure
• Archimedes’ Principle
• Continuity Equation
• Bernoulli’s Equation
• Viscosity and Viscous Drag
• Surface Tension

Pressure
Pressure arises from the collisions between the particles of a fluid
with another object (container walls for example).
There is a momentum
change (impulse) that is
away from the container
walls. There must be a
force exerted on the
particle by the wall.

Pressure is defined as .
A
F
P 
The units of pressure are N/m2 and are called Pascals
(Pa).
Note: 1 atmosphere (atm) = 101.3 kPa
By Newton’s 3rd Law, there is a force on the wall due
to the particle.

Example (text problem 9.1): Someone steps on your toe, exerting
a force of 500 N on an area of 1.0 cm2. What is the average
pressure on that area in atmospheres?
atm
49
Pa
10
013
.
1
atm
1
N/m
1
Pa
1
N/m
10
0
.
5
m
10
1.0
N
500
5
2
2
6
2
4
av


















 
A
F
P
2
4
2
2
m
10
0
.
1
cm
100
m
1
cm
0
.
1 








A 500N person
weighs about
113 lbs.

Gravity’s Effect on Fluid Pressure
An imaginary
cylinder of
fluid
FBD for the fluid cylinder
P1A
P2A
w
x
y
Imaginary cylinder
can be any size

Apply Newton’s 2nd Law to the fluid cylinder. Since the
fluids isn’t moving the net force is zero.
 
gd
P
P
gd
P
P
gd
P
P
g
Ad
A
P
A
P
w
A
P
A
P
F




















1
2
1
2
1
2
1
2
1
2
or
0
0
0
If P1 (the pressure at the top of the cylinder) is known, then the
above expression can be used to find the variation of pressure
with depth in a fluid.

If the top of the fluid column is placed at the surface of the fluid,
then P1 = Patm if the container is open.
gd
P
P 

 atm
You noticed on the previous slide that the areas canceled out.
Only the height matters since that is the direction of gravity.
Think of the pressure as a force density in N/m2

Example (text problem 9.19): At the surface of a freshwater lake,
the pressure is 105 kPa. (a) What is the pressure increase in going
35.0 m below the surface?
   
atm
3.4
kPa
343
m
35
m/s
8
.
9
kg/m
1000 2
3
atm
atm









gd
P
P
P
gd
P
P



Example: The surface pressure on the planet Venus is 95 atm.
How far below the surface of the ocean on Earth do you need to be
to experience the same pressure? The density of seawater is 1025
kg/m3.
  
m
950
N/m
10
9.5
m/s
8
.
9
kg/m
1025
N/m
10
9.5
atm
94
atm
1
atm
95
2
6
2
3
2
6
atm










d
d
gd
gd
gd
P
P




Measuring Pressure
A manometer is a
U-shaped tube that
is partially filled
with liquid,
usually Mercury
(Hg).
Both ends of the
tube are open to the
atmosphere.

A container of gas is connected to one end of the U-tube
If there is a pressure difference between the gas and the atmosphere, a force
will be exerted on the fluid in the U-tube. This changes the equilibrium
position of the fluid in the tube.

Also
atm
c P
P 
At point C
B'
B P
P 
The pressure at point B is the pressure of the gas.
gd
P
gd
P
P
P
P
gd
P
P
P
B
C
B
C
B
B











gauge
atm
'
From the figure:
gauge meas atm
P = P - P

A Barometer
The atmosphere pushes on the container of mercury which forces
mercury up the closed, inverted tube. The distance d is called the
barometric pressure.

Atmospheric pressure is equivalent to a column of
mercury 76.0 cm tall.
gd
PA 

From the figure atm
B
A P
P
P 

and

The Many Units of Pressure
1 ATM equals 1.013x105 N/m2
14.7 lbs/in2
1.013 bar
76 cm Hg
760 mm Hg
760 Torr
34 ft H2O
29.9 in Hg

Pascal’s Principle
A change in pressure at any point in a confined fluid is
transmitted everywhere throughout the fluid. (This is
useful in making a hydraulic lift.)

Apply a force F1 here
to a piston of cross-
sectional area A1.
The applied force is
transmitted to the piston
of cross-sectional area
A2 here.
In these problems neglect
pressure due to columns
of fluid.

Mathematically,
1
1
2
2
2
2
1
1
A
A
A
F
A
F
2
point
at
1
point
at
F
F
P
P














Example: Assume that a force of 500 N (about 110 lbs) is applied to
the smaller piston in the previous figure. For each case, compute
the force on the larger piston if the ratio of the piston areas (A2/A1)
are 1, 10, and 100.
50,000 N
100
5000 N
10
500 N
1
F2
1
2 A
A
Using Pascal’s Principle:

Archimedes’ Principle
An FBD for an object floating
submerged in a fluid.
The total force on the block due to
the fluid is called the buoyant force.
1
2
1
2
where F
F
F
F
F



B
w
F2
F1
x
y

The magnitude of the buoyant force is:
 A
P
P
A
P
A
P
F
F
FB
1
2
1
2
1
2






gd
P
P 

 1
2
From before:
gV
gdA
FB 
 

The result is
Buoyant force = the weight of the fluid displaced

gV
FB 

Archimedes’ Principle: A fluid exerts an upward buoyant force
on a submerged object equal in magnitude to the weight of the
volume of fluid displaced by the object.

Example (text problem 9.28): A flat-bottomed barge loaded with
coal has a mass of 3.0105 kg. The barge is 20.0 m long and 10.0 m
wide. It floats in fresh water. What is the depth of the barge below
the waterline?
x
y
w
FB
FBD for
the barge
Apply Newton’s 2nd Law to the barge:
 
  b
w
b
w
w
b
w
w
w
B
B
m
Ad
m
V
g
m
g
V
g
m
w
F
w
F
F












0
  
m
5
.
1
m
10.0
*
m
0
.
20
kg/m
1000
kg
10
0
.
3
3
5




A
m
d
w
b


Example (text problem 9.40): A piece of metal is released under
water. The volume of the metal is 50.0 cm3 and its specific gravity is
5.0. What is its initial acceleration? (Note: when v = 0, there is no
drag force.)
FBD for
the metal ma
w
F
F B 



Vg
FB water


The magnitude of the buoyant force
equals the weight of the fluid displaced
by the metal.
Solve for a: 












 1
ρ
ρ
ρ
ρ
object
object
water
object
object
water
V
V
g
g
V
Vg
g
m
F
a B
Apply Newton’s 2nd Law to
the piece of metal:
x
y
w
FB

Since the object is completely submerged V=Vobject.
water
gravity
specific



where water = 1000 kg/m3 is the
density of water at 4 °C.
Given 0
.
5
gravity
specific
water
object




2
object
object
water
m/s
8
.
7
1
0
.
5
1
1
.
.
1
1
ρ
ρ



























 g
G
S
g
V
V
g
a
Example continued:
The sign is minus because gravity acts down. BF causes a < g.

Fluid Flow
A moving fluid will exert forces parallel to the surface over which
it moves, unlike a static fluid. This gives rise to a viscous force
that impedes the forward motion of the fluid.
A steady flow is one where the velocity at a given point in a
fluid is constant.
V1 =
constant
V2 =
constant
v1v2

Steady flow is laminar; the fluid flows in layers. The
path that the fluid in these layers takes is called a
streamline.
An ideal fluid is incompressible, undergoes laminar
flow, and has no viscosity.
Streamlines do not cross.
Crossing streamlines would indicate a volume of fluid with
two different velocities at the same time.

The Continuity Equation—Conservation of Mass
The amount of mass that flows though the cross-sectional area A1
is the same as the mass that flows through cross-sectional area A2.
Faster Slower

is the mass flow rate (units kg/s)
Av
t
m




Av
t
V


 is the volume flow rate (units m3/s)
2
2
2
1
1
1 v
A
v
A 
 
The continuity equation is
If the fluid is incompressible, then 1= 2.

Example (text problem 9.41): A garden hose of inner radius 1.0 cm
carries water at 2.0 m/s. The nozzle at the end has radius 0.20 cm.
How fast does the water move through the constriction?
  m/s
50
m/s
0
.
2
cm
0.20
cm
0
.
1
2
1
2
2
2
1
1
2
1
2
2
2
1
1



























v
r
r
v
A
A
v
v
A
v
A


Simple ratios

Bernoulli’s Equation
Bernoulli’s equation is a statement of energy
conservation.

2
2
2
2
2
1
1
1
2
1
2
1
v
gy
P
v
gy
P 


 




Potential
energy
per unit
volume
Kinetic
energy
per unit
volume
Work per unit
volume done
by the fluid
Points 1 and 2
must be on the
same streamline
This is the most general equation

Example (text problem 9.49): A nozzle is connected to a horizontal
hose. The nozzle shoots out water moving at 25 m/s. What is the
gauge pressure of the water in the hose? Neglect viscosity and
assume that the diameter of the nozzle is much smaller than the
inner diameter of the hose.
2
2
2
2
2
1
1
1
2
1
2
1
v
gy
P
v
gy
P 


 




Let point 1 be inside the hose and point 2 be outside the nozzle.
The hose is horizontal so y1 = y2. Also P2 = Patm.

2
1
2
2
atm
1
2
2
atm
2
1
1
2
1
2
1
2
1
2
1
v
v
P
P
v
P
v
P










Substituting:
v2 = 25 m/s and v1 is unknown. Use the continuity equation.
2
2
1
2
2
2
1
2
2
2
1
2
1
2
2
v
d
d
v
d
d
v
A
A
v 














































Since d2<<d1 it is true that v1<<v2.
Example continued:

 
  
Pa
10
1
.
3
m/s
25
kg/m
1000
2
1
2
1
2
1
2
1
2
1
5
2
3
2
2
2
1
2
2
2
1
2
2
atm
1









v
v
v
v
v
P
P




Example continued:

1
Since v 0

Viscosity
A real fluid has viscosity (fluid friction). This implies a
pressure difference needs to be maintained across the
ends of a pipe for fluid to flow.

Viscosity also causes the existence of a velocity gradient across a
pipe. A fluid flows more rapidly in the center of the pipe and
more slowly closer to the walls of the pipe.
The volume flow rate for laminar flow of a viscous fluid is given
by Poiseuille’s Law.
4
8
r
L
P
t
V

 



where  is the viscosity
4th power

Example (text problem 9.55): A hypodermic syringe attached to a
needle has an internal radius of 0.300 mm and a length of 3.00 cm.
The needle is filled with a solution of viscosity 2.0010-3 Pa sec; it is
injected into a vein at a gauge pressure of 16.0 mm Hg. Neglect the
extra pressure required to accelerate the fluid from the syringe into the
entrance needle.
(a) What must the pressure of the fluid in the syringe be in
order to inject the solution at a rate of 0.250 mL/sec?
Solve Poiseuille’s Law for the pressure difference:
  
 
Pa
4716
sec
cm
250
.
0
cm
10
3
.
0
cm
00
.
3
sec
Pa
10
00
.
2
8
8 3
4
1
3
4














t
V
r
L
P

Example continued:
This pressure difference is between the fluid in the syringe
and the fluid in the vein. The pressure in the syringe is
Pa
6860
Pa
4720
Pa
2140 








P
P
P
P
P
P
v
s
v
s
5
1.013x10 Pa
16 (mm Hg)× = 2132 Pa
760 mm Hg
Conversion:

(b) What force must be applied to the plunger, which
has an area of 1.00 cm2?
Example continued:
The result of (a) gives the force per unit area on
the plunger so the force is just F = PA = 0.686 N.

Viscous Drag
The viscous drag force on a sphere is given by Stokes’ law.
rv
FD 
6

Where  is the viscosity of the fluid that the sphere is falling
through, r is the radius of the sphere, and v is the velocity of
the sphere.

Example (text problem 9.62): A sphere of radius 1.0 cm is dropped
into a glass cylinder filled with a viscous liquid. The mass of the
sphere is 12.0 g and the density of the liquid is 1200 kg/m3. The
sphere reaches a terminal speed of 0.15 m/s. What is the viscosity
of the liquid?
FBD for
sphere ma
w
F
F
F B
D 




FD
w
x
y
FB
Apply Newton’s Second Law to
the sphere
Drag Buoyant

Example continued:
When v = vterminal , a = 0 and
sec
Pa
4
.
2
6
0
6
0
6
0
6
0















t
s
l
s
s
s
l
t
s
l
l
t
s
l
t
B
D
rv
g
V
g
m
g
m
g
V
rv
g
m
g
V
rv
g
m
g
m
rv
w
F
F








Solving for 

Surface Tension
The surface of a fluid acts like a stretched membrane
(imagine standing on a trampoline). There is a force
along the surface of the fluid.
The surface tension is a force per unit length.

Example (text problem 9.70): Assume a water strider has a roughly
circular foot of radius 0.02 mm. The water strider has 6 legs.
(a) What is the maximum possible upward force on the foot
due to the surface tension of the water?
The water strider will be able to walk on water if the
net upward force exerted by the water equals the
weight of the insect. The upward force is supplied by
the water’s surface tension.
N
10
9
2 6
2 









 r
r
PA
F 


(b) What is the maximum mass of this water strider so that it
can keep from breaking through the water surface?
Example continued:
To be in equilibrium, each leg must support one-
sixth the weight of the insect.
kg
10
5
6
or
6
1 6





g
F
m
w
F

Summary
•Pressure and its Variation with Depth
•Pascal’s Principle
•Archimedes Principle
•Continuity Equation (conservation of mass)
•Bernoulli’s Equation (conservation of energy)
•Viscosity and Viscous Drag
•Surface Tension

Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?
1. Rise
2. Fall
3. Remain unchanged

1. Rise
2. Fall
3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is
thrown overboard into the pool, will the water level at the edge of the
pool rise, fall, or remain unchanged?

In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball
1. rises.
2. falls.
3. remains in place.

The weight of the stand and suspended solid iron ball is equal to the weight
of the container of water as shown above. When the ball is lowered into the
water the balance is upset. The amount of weight that must be added to the
left side to restore balance, compared to the weight of water displaced by
the ball, would be
1. half. 2. the same.
3. twice. 4. more than twice.

A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is
1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.

Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will
1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.

The density of the block of
wood floating in water is
1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.

1. greater than the density of water.
2. equal to the density of water.
3. less than half that of water.
4. more than half the density of water.
5. … not enough information is given.
The density of the block of
wood floating in water is

1. rises.
2. falls.
3. remains in place.
In the presence of air, the small iron ball and large plastic ball balance
each other. When air is evacuated from the container, the larger ball

1. half. 2. the same.
3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the weight
of the container of water as shown above. When the ball is lowered into the
water the balance is upset. The amount of weight that must be added to the
left side to restore balance, compared to the weight of water displaced by
the ball, would be

1. upset and the stick rotates clockwise.
2. upset and the stick rotates counter-clockwise.
3. unchanged.
A pair of identical
balloons are inflated
with air and
suspended on the
ends of a stick
that is horizontally
balanced. When the
balloon on the left is punctured,
the balance of the stick is

1. sink.
2. return to the surface.
3. stay at the depth to
which it is pushed.
Consider an air-filled balloon weighted so that it is on the verge of
sinking—that is, its overall density just equals that of water.
Now if you push it beneath the surface, it will

Basics of Fluids

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