This document discusses fluids and fluid mechanics. It defines a fluid as anything that flows, including liquids and gases. It discusses the properties of fluids like density, pressure, viscosity, compressibility, and how these properties depend on factors like temperature. It introduces concepts like Pascal's principle, Archimedes' principle, Bernoulli's principle, and equations like the equation of continuity that relate key variables in fluid flow situations. Examples are provided to illustrate how to apply these principles and equations to calculate things like fluid pressure, velocity, and buoyant forces.

1. Fluids

2. solid liquid
gas
Fluid Mechanics
•A fluid is anything that flows:
•Fluids can be a liquid or a gas
•e.g. liquid (water, mercury at room temp) gas (steam, air, CO2)
•Properties depend on temperature

3. Fluids
A container is needed to keep gas in a
certain location
A solid has a rigid structure due to
strong bonding
A liquid has weak bonds
A gas is free to move – molecules
independent

4. Fluids - Density
The density ρ is the mass m of a
substance divided by its volume V. {the
density is defined as the mass of a unit
volume}
SI units of density is kgm-3
So equal volumes of substances
generally have different masses and
therefore different densities.
V
m
=ρ

5. Mass Density
Equal volumes of different substances
generally have different masses. Characterise
using density (ρ)
V
m
volume
mass
==ρ
Material Density (kg /m3
)
Aluminium 2700
Gold 19300
Ice 917
Water 1000
Oil 800
Air 1.29
Helium 0.179

6. Pressure (P)
Pressure is defined as the force acting
perpendicular to a unit area. In other
words, it is given by perpendicular
force (F) to the surface divided by the
area (A) over which the force acts.
SI unit of pressure:1 N/m2
= 1 Pa or 1
Pascal.
Atmospheric pressure ≈ 105
Pa
A
F
P =
© John Wiley and Sons, 2004

7. 1. A cylinder of 4 cm radius is filled with water
to the height of 10 cm. find the pressure at
the bottom of the cylinder.
2. Cylindrical shape tube is filled with water.
The pressure at the bottom is measured as
1 atm. Find the height of the water level.
3. If it is filled with mercury, find the height of
the mercury level.
(Density of Hg-13600 kgm-3
Water-1000 kgm-3
)
Examples:

8. Pressure with Depth in a static fluid
Pressure in a
static fluid is
proportional to
depth.
© John Wiley and Sons, 2004

9. Pressure with depth
Take a column of liquid in equilibrium
Sum of forces is zero.
Relative to surface P1 = 0 so
ghPP
ghAAPAPF
ghAm
AhV
Vm
mgAPAPF
y
y
ρ
ρ
ρ
ρ
+=
=−−=
=
=
=
=−−=
∑
∑
12
12
12
0
0
ghP ρ=
(÷A)

10. Examples:
1. Find the pressure on an object of 10m depth at sea.
(density of sea water is given as 1030 kgm-3
and
the atmospheric pressure is given as 105
Pa)
2. Find the pressure under 1m of,
a. Water
b. Oil(density is 800 kgm-3
)
c. Hg(density – 13600kgm-3
)
3 At what depth in the water the atmospheric pressure
is doubled

11. Pascal’s Principle
Any change in pressure applied to a completely enclosed fluid is
transmitted undiminished to all parts of the fluid and the closed
walls.
By applying force to a small area (F1) this pressure can be used
in hydraulic systems to lift large masses e.g. car lift






=
=
=
1
2
12
1122
A
A
FF
AFAF
A
F
P

12. Car lift

13. Problem:
1. If a downward force F1 = 50 N is applied to the
small piston with area A1 = 100 cm2
, what
upward force F2 does the fluid apply to the
large piston with area A2 = 1000 cm2
?
Remember: Pressure is the same everywhere.

14. Problem 1
A barber raises his customer’s chair by applying a
force of 150N to a hydraulic piston of area 0.01 m2
. If
the chair is attached to a piston of area 0.1 m2
, how
massive a customer can the chair raise? Assume the
chair itself has a mass of 5 kg.

15. Answer
To solve this problem, first determine the force applied to the larger piston.
If the maximum force on the chair is 1500N, you can now determine the
maximum mass which can be lifted by recognizing that the force that must be
overcome to lift the customer is the force of gravity, therefore the applied force
on the customer must equal the force of gravity on the customer.
If the chair has a mass of 5 kilograms, the maximum mass of a customer in the
chair must be 148 kg.

16. Problem 2
A hydraulic system is used to lift a 2000-kg
vehicle in an auto garage. If the vehicle sits
on a piston of area 1 square meter, and a
force is applied to a piston of area 0.03
square meters, what is the minimum force
that must be applied to lift the vehicle?

17. Answer

18. Archimedes’ Principle
When a body is wholly or partially immersed in a fluid
there is an up thrust which is equal to the weight of
fluid displaced.

19. Archimedes’ Principle
P2>P1 so the upward force exceeds the
downward force!
The buoyant force FB = weight of the
liquid displaced
( )
displacedliquidofweightghA
Vm
densityliquid
volumehA
ghAAPPAPAPFB
=
=
=
=
=−=−=
ρ
ρ
ρ
ρ1212

20. 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: so we can express that by the
following equations.
If the buoyant force is strong enough to displace the
force of gravity then the object will float in the fluid
fluidB WF = gmgV fluidobjimmersed =ρ

21. Archimedes’ Principle
An object of weight 100N
is immersed in water. The
deeper the object is the
more fluid it displaces and
the stronger the buoyancy
force is. In (b) the buoyant
force matches the weight
of the cube so it floats!
fluidB WF =
© John Wiley and Sons, 2004

22. Archimedes’ Principle
A solid square pine raft measures 4m by 4m by 0.3m
thick. Determine if the raft will float in water if the
density of the wood is 550kg/m3
. Density of water is
1000kg/m3
.
© John Wiley and Sons, 2004

23. Archimedes’ Principle
In the last example – compare the maximum possible
buoyant force and the weight of the raft.
The comparison depends on the densities only!
Any object solid throughout will float if the density of
the material is less than the liquid
What about a ship made of metal?
It is not solid- hollow.. but has a large area to
displace lots of water to balance its weight.
( ) ( )gVgV woodwaterwoodwood ρρ ⇔

24. Archimedes' Principle (Cont.)
Does Archimedes'
principle depend on
depth?
 No. Volume of fluid
displaced by
submerged object is
the same at any
depth.
 Buoyant force only
depends on pressure
differences
 Same at any depth

25. Sink or Float
An object more (less) dense than
the fluid in which it is immersed
will sink (float).
What must you do to increase
your buoyancy in water?

26. Sink or Float
An object more (less) dense
than the fluid in which it is
immersed will sink (float).
 What must you do to increase
your buoyancy in water?
 Decrease your weight density
 Life jacket increases volume
while adding very little weight
An object with a density equal
to the surrounding fluid will
neither sink nor float.
 Fish (air sac) and submarines
(ballast) can vary their densities
Crocodiles increase their
weight
density by swallowing stones.

27. The Principle of Flotation
How much water must
be displaced for an
object to float?
Principle of flotation
 A floating object
displaces a volume of
fluid with a weight equal
to its own weight.

28. Archimedes figured out that the meta center
had to be determined which is a point where
an imaginary vertical line (through the center of
buoyancy) intersects another imaginary vertical
line (through a new centre of buoyancy)
created after the ship is displaced, or tilted, in
the water.
How is it that an iron ship will float in water?

29. The center of buoyancy in a floating ship is the point in which all the
body parts exactly balance each other and make each other float.
In other words, the meta center remains directly above the center of
buoyancy regardless of the tilt of the floating ship.
When a ship tilts, one side displaces more water than the other
side, and the center of buoyancy moves and is no longer directly
under the center of gravity; but regardless of the amount of the tilt,
the center of buoyancy remains directly below the meta center.
If the meta center is above the center of gravity, buoyancy restores
stability when the ship tilts. If the meta center is below the center of
gravity, the boat is unstable and capsizes.

30. Archimedes’ Principle

31. Problem 1.
1. A body weighs 450 g in air and 310 g
when completely immersed in water.
Find
(i) the loss in weight of the body
(ii) the upthrust on the body
(iii) the volume of the body

32. Answer 1
(i) loss in weight = weight in air - weight in water
= 450 - 310
= 140gf.
(ii) upthrust = loss of weight
= 140 gf
(iii) weight of displaced water = upthrust = 140gf
mass of displaced water = 140g
Assumption: density of water is 1 gcm-3
Volume of displaced water =140/1=140cm3
Volume of the body = volume of displaced water = 140 cm3

33. 2. The mass of a block made of certain
material is 13.5 kg and its volume is 15
x 10-3
m3
. Will the block float or sink in
water? Give a reason for your answer.

34. Density of the block = 0.9 x 103
kg m-3
(density of water is 1 x 103
kg m-3
)
The block floats in water because its
density is less than that of water.

35. 3. A hollow cubical box is 0.30 meters on
an edge. This box is floating in a lake
with 1/3 of its height beneath the
surface. The walls of the box have a
negligible thickness. Water is poured
into the box. What is the depth of the
water in the box at the instant the box
begins to sink?

36. Fluids in motion
Flowing water or even wind is a
moving fluid. It is important to be able
to handle such phenomena.
We use a few parameters to
describe the motion of the fluid
particles
Steady or Unsteady flow
(turbulence)
Compressible or Incompressible
Viscous or Non-viscous

37. Steady/Unsteady flow
In steady flow the velocity of particles is constant with
time
Unsteady flow occurs when the velocity at a point
changes with time. Turbulence is extreme unsteady
flow where the velocity vector at a point changes
quickly with time e.g. water rapids or waterfall.
When the flow is steady, streamlines are used to
represent the direction of flow.
Steady flow is sometimes called streamline flow

38. Compressible/Incompressible
fluids
Incompressible fluid the density remains constant with
changing pressure.
Most liquids can be modelled as incompressible
Some gasses cannot be said to be incompressible,
therefore they are compressible.

39. Viscous/Non viscous flow
A viscous fluid such as honey does not flow readily, it
has a large viscosity.
Water has a low viscosity and flows easily.
A viscous flow requires energy dissipation.
Zero viscosity – requires no energy. Some liquids can
be taken to have zero viscosity e.g. water.
An incompressible, nonviscous fluid is said to be an
ideal fluid

40. Equation of continuity
The mass flow rate
(ρAv)has the same
value at every point
along a tube that
has a single entry
and exit point for the
fluid flow. For two
positions
222111 vAvA ρρ =
If the area is
reduced the fluid
must speed up!

41. Equation of continuity
The mass flow rate
(ρAv) unit is kg/s






==





∆
∆
secsec1sec
2
3222
kgmm
m
kg
vA
kg
t
m
ρ
If the area is
reduced the fluid
must speed up!

42. Equation of continuity
If the fluid is incompressible, the density
remains constant throughout
Av represents the volume of fluid per
second that passes through the tube
and is called the volume flow rate Q
2211 vAvA =
AvQ =

43. Problem 1
Water runs through a water main of
cross-sectional area 0.4 m2
with a
velocity of 6 m/s. Calculate the velocity
of the water in the pipe when the pipe
tapers down to a cross-sectional area
of 0.3 m2
.

44. Answer

45. Problem 2
Water enters a typical garden hose of
diameter 1.6 cm with a velocity of 3 m/s.
Calculate the exit velocity of water from the
garden hose when a nozzle of diameter 0.5
cm is attached to the end of the hose.

46. Answer
First, find the cross-sectional areas of the entry
(A1) and exit (A2) sides of the hose.
Next, apply the continuity equation for fluids to
solve for the water velocity as it exits the hose
(v2).

47. Bernoulli’s Principle
In steady flow, the speed, pressure and elevation of an
ideal fluid are related. Two situations.
(1) Horizontal pipe, changing Area A.
 Pressure drop in thin pipe why?
 Fluid speeds – acceleration – force required
(2) Elevation change but same area A
 Lower fluid is under more pressure

48. Bernoulli’s Principle
In steady flow for a non-viscous incompressible fluid
the density ρ, the pressure P, the fluid velocity v and
elevation y are related through
2
2
22
1
21
2
12
1
1 gyvPgyvP ρρρρ ++=++

49. Examples of Bernoulli’s Principle
(a) good plumbing, (b) bad plumbing
In (a) pressure difference between points A and B, water trap
flushed out – sewer gas in house
In (b) no pressure difference between points A and B using
vent, water trap kept in place – no sewer gas in house

50. Examples of Bernoulli’s Principle
Air moves faster over top of wing, slower
beneath. Leads to lower pressure over wing
and uplift to raise plane!

51. Examples of Bernoulli’s Principle
Curve ball, air on one side of spinning ball
faster than the other (leads to pressure
difference!)

52. Example 1
Water circulates throughout a house in a hot-
water heating system. If water is pumped out
at a speed of 0.50 m/s through a 4.0 cm
diameter pipe in the basement under a
pressure of 3.0 atm, what will be the flow
speed and pressure in a 2.6 cm diameter
pipe on the second floor 5.0 m above?
Assume pipes do not divide into branches.

53. Answer 1
First calculate the flow speed on the second floor,
calling it v2, using the equation of continuity. We can
call the basement point 1.
v2 = (v1A1) / (A2)
= (v1π(r1)2
) / (π(r2)2
)
= (0.50 m/s)(0.020 m)2
/ (0.013 m)2
= 1.2 m/s

54. To find pressure, we use Bernoulli’s equation:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2
– (v2)2
)
P2 = (3.0 x 105
N/m2
) + (1.0 x 103
kg/m3
) [ (0.50 m/s)2
– (1.2 m/s)2
]
P2 = (3.0 x 105
N/m2
) – (4.9 x 104
N/m2
) – (6.0 x 102
N/m2
)
P2 = 2.5 x 105
N/m2

55. Problem 1
What is the lift (in Newtons) due to
Bernoulli’s principle on a wing of area
86 m2 if the air passes over the top and
bottom surfaces at speeds of 340 m/s
and 290 m/s, respectively?

56. Problem 2
Water at a gauge pressure of 3.8 atm at
street level flows in to an office building
at a speed of 0.06 m/s through a pipe
5.0 cm in diameter. The pipes taper
down to 2.6cm in diameter by the top
floor, 20 m above. Calculate the flow
velocity and the gauge pressure in such
a pipe on the top floor. Assume no
branch pipe and ignore viscosity.

57. Problem 3
The water supply of a building is fed
through a main entrance pipe 6.0 cm in
diameter. A 2.0 cm diameter faucet tap
positioned 2.0 m above the main pipe
fills a 25 liter container in 30 s.
(a) What is the speed at which the
water leaves the faucet.
(b) What is the gauge pressure in the
main pipe.
Assume that the faucet is the only outlet
in the system