01 The Nature of Fluids
(Water Resources I)
Dave Morgan
AT X 2ε ,
Prepared using Lyx, and the Beamer class in L
E
on September 12, 2007
01 The Nature of Fluids WRI 1/17
Recommended Text
A recommended text to accompany these notes is Applied Fluid
Mechanics by Mott:
Read sections:
1.3, 1.4, 1.6 (omit US units), 1.7, 1.8, 1.9, 1.11
Study Example Problems: 1.5 - 1.9
01 The Nature of Fluids WRI 2/17
Elementary Properties of Fluids
Fluids can be either liquid or gas
A liquid tends to ow and conform to the shape of its
container
Liquids are not readily compressible (for the purpose of this
course, we consider them to be imcompressible)
A gas tends to expand to ll the closed container it is in (or to
disperse if not contained).
Gases are readily compressible
We shall be primarily concerned with liquids
01 The Nature of Fluids WRI 3/17
Primary Units
SI units are used. Four primary units will be used extensively in this
course:
Quantity
SI unit
Dimension
Length
metre, m
L
Mass
kilogram, kg
M
Time
second, s
T
Temperature
Kelvin, K
θ
(Current and luminosity are the two other primary SI units.)
01 The Nature of Fluids WRI 4/17
Derived Units
Quantity
SI Unit
velocity
m/s
LT
−1
2
LT
−2
acceleration
m/s
force
N, newton
energy
J, joule
(work)
N·m
2
kg·m/s
kg m
power
2 /s2
N · m/s
Dimensions
MLT
−2
2 −2
ML T
2 −3
ML T
J/s
pressure
(stress)
Pa, pascal
2
N/m
kg/m/s
ML
−1 T−2
2
01 The Nature of Fluids WRI 5/17
Derived Units:
Quantity
SI Unit
Volume ow rate,
Q
3
m /s
Dimensions
3 −1
L T
L/s
Weight ow rate,
Mass ow rate,
Specic weight,
Density,
ρ
W
N/s
2
kg/m/s
M
kg/s
γ
N/m
3
kg/m
3
−1 S−2
ML
MT
−1
−2 T−2
ML
ML
−3
01 The Nature of Fluids WRI 6/17
Pressure
Pressure is given by:
p=
F
A
It is the force per unit area on a surface, where
2
1 N/m
= 1 Pa
(pascal)
01 The Nature of Fluids WRI 7/17
Pressure
Pressure is given by:
p=
F
A
It is the force per unit area on a surface, where
2
1 N/m
= 1 Pa
(pascal)
Blaise Pascal (1623 - 1662), after whom the Pascal programming
language was named, determined the following principles:
1 Pressure acts uniformly in all
directions on a small volume of a
uid at rest
2 In a uid conned by solid
boundaries, pressure acts
perpendicularly to the boundaries
01 The Nature of Fluids WRI 7/17
Pascal's Laws
Pressure acts uniformly in all directions
on a small volume of a uid at rest.
The forces must balance out
(i.e.
Σ Fx = Σ Fy = 0);
otherwise the
volume of uid will not be in equilibrium
and cannot remain at rest.
Also, the volume must be suciently
small that we do not have to consider
the mass, and therefore the weight, of
the volume of uid. If the weight is not
negligible, the upward pressure on the
bottom of the volume will have to be
greater than the downward pressure on
the top of the volume so that
Σ Fy = Fup − W − Fdown = 0.
01 The Nature of Fluids WRI 8/17
Pascal's Laws
In a uid conned by solid
boundaries, pressure acts
perpendicularly to the boundaries.
Why is this true?
(Consider a small volume of uid at
rest against one of the boundaries?
If this volume remains at rest, what
are the forces that act upon it?)
01 The Nature of Fluids WRI 9/17
Density
Density is mass per unit volume:
ρ=
m
V
01 The Nature of Fluids WRI 10/17
Density
Density is mass per unit volume:
ρ=
◦
m
V
◦
The density of water between 0 C and 15 C is close to
3
1000 kg/m .
◦
It has a maximum density at 4 C.
◦
3
Above 15 C, the density drops steadily to a density of 958 kg/m
◦
at 100 C.
(There is a table of values for the properties of water at the back of
Applied Fluid Mechanics by Mott, or from numerous other sources)
01 The Nature of Fluids WRI 10/17
Specic Weight
Specic weight is weight per unit volume:
γ=
w
V
01 The Nature of Fluids WRI 11/17
Specic Weight
Specic weight is weight per unit volume:
γ=
w
V
3 between 0 ◦ C and 15 ◦ C.
Water has a specic weight of 9.81 kN/m
01 The Nature of Fluids WRI 11/17
Specic Weight
Specic weight is weight per unit volume:
γ=
w
V
3 between 0 ◦ C and 15 ◦ C.
Water has a specic weight of 9.81 kN/m
Since
w = mg ,
it follows that:
γ=
w mg
=
= ρg
V
V
01 The Nature of Fluids WRI 11/17
Specic Gravity
Specic gravity is the ratio of the density (or specic weight) of a
◦
substance to the density (or specic weight) of water at 4 C.
Then, the specic gravity of a substance
sg =
γs
γw @ 4 ◦ C
=
s
is given by
ρs
ρw @4 ◦ C
01 The Nature of Fluids WRI 12/17
Specic Gravity
Specic gravity is the ratio of the density (or specic weight) of a
◦
substance to the density (or specic weight) of water at 4 C.
Then, the specic gravity of a substance
γs
sg =
γw @ 4 ◦ C
=
s
is given by
ρs
ρw @4 ◦ C
◦
3 and the density of
The density of gasoline at 25 C is 680 kg/m
◦
3 . Therefore, the specic gravity of
water at 4 C is 1000 kg/m
◦
gasoline at 25 C is
sg = 680/1000 = 0.68.
01 The Nature of Fluids WRI 12/17
Specic Gravity
Specic gravity is the ratio of the density (or specic weight) of a
◦
substance to the density (or specic weight) of water at 4 C.
Then, the specic gravity of a substance
γs
sg =
γw @ 4 ◦ C
=
s
is given by
ρs
ρw @4 ◦ C
3 and the density of
◦
The density of gasoline at 25 C is 680 kg/m
◦
3 . Therefore, the specic gravity of
water at 4 C is 1000 kg/m
◦
gasoline at 25 C is
sg = 680/1000 = 0.68.
◦
The specic weight of mercury at 25 C is 132.8 kN/m
3 and the
◦
3
specic weight of water at 4 C is 9.81 kN/m so the specic
◦
gravity of mercury at 25 C is sg = 13.54
01 The Nature of Fluids WRI 12/17
Nature of Fluids
Example
Calculate the pressure produced in the oil in a closed cylinder by a
piston with diameter 7.5 cm exerting a force of 11 175 N
01 The Nature of Fluids WRI 13/17
Nature of Fluids
Example
Calculate the pressure produced in the oil in a closed cylinder by a
piston with diameter 7.5 cm exerting a force of 11 175 N
11175 N
7.5 cm
01 The Nature of Fluids WRI 13/17
Nature of Fluids
Example
Calculate the pressure produced in the oil in a closed cylinder by a
piston with diameter 7.5 cm exerting a force of 11 175 N
11175 N
Solution
p =
=
F
A
11175 N
π(0.075)2 /4 m2
=
2529500 Pa
=
2.53 MPa
7.5 cm
01 The Nature of Fluids WRI 13/17
Nature of Fluids
Example
Calculate the weight of 1 m
3 of kerosene if it has a mass of 823 kg
01 The Nature of Fluids WRI 14/17
Nature of Fluids
Example
Calculate the weight of 1 m
3 of kerosene if it has a mass of 823 kg
Solution
W
Note:
= mg
2
=
823 kg × 9.81 m/s
=
8070 N
In general, use 5 signicant gures for interim calculations
and 3 signicant gures for displayed solutions.
01 The Nature of Fluids WRI 14/17
Nature of Fluids
Example
Calculate the density and the specic weight of benzene if its
specic gravity is 0.876.
01 The Nature of Fluids WRI 15/17
Nature of Fluids
Example
Calculate the density and the specic weight of benzene if its
specic gravity is 0.876.
Solution
ρb
ρwater @4◦ C
ρb = 0.876 × 1000 kg/m3
= 876 kg/m3
0.876
=
γb
γwater @4◦ C
γb = 0.876 × 9.81 kN/m3
= 8.59 kN/m3
0.876
=
01 The Nature of Fluids WRI 15/17
Nature of Fluids
Example
◦
A cylindrical tank with diameter 12.0 m contains water at 20 C to
◦
a depth of 4.0 m. If the water is heated to 65 C, what is the depth
of the water? (Assume that the tank dimensions remain constant
and that there are no losses due to evaporation.)
01 The Nature of Fluids WRI 16/17
Nature of Fluids
Example
◦
A cylindrical tank with diameter 12.0 m contains water at 20 C to
◦
a depth of 4.0 m. If the water is heated to 65 C, what is the depth
of the water? (Assume that the tank dimensions remain constant
and that there are no losses due to evaporation.)
Solution
◦
Volume at 20 C:
V20 =
π d 2 h20
4
=
π(12.0 m)2 (4.0 m)
4
= 452.39 m3
Mass of water in the tank:
m = ρ V20 = 998 kg/m3 × 452.39 m3 = 451490 kg
...continued
01 The Nature of Fluids WRI 16/17
Nature of Fluids
Solution (continued)
V20 =
π d 2 h20
4
=
π(12.0 m)2 (4.0 m)
4
= 452.39 m3
m = ρ20 V20 = 998 kg/m3 × 452.39 m3 = 451490 kg
◦
Volume at 65 C:
V65 =
m
ρ65
=
451490 kg
981 kg/m3
= 460.23 m3
◦
Depth at 65 C:
h65 =
3
V
4 × 460.23 m
=
= 4.0639 m
πd2
π(12.0)2
4 65
◦
The depth at 65 C is 4.06 m
01 The Nature of Fluids WRI 17/17