Introduction to Fluid
Mechanics
1
“Anything that
flows”
Definition of Fluid
It is a substance that deforms continuously when acted on by shearing stress of any magnitude.
A shearing stress (force per unit area) is created whenever a tangential force acts on a surface
U
𝝉𝝉 = −μ
𝑑𝑑(𝑣𝑣)
𝑑𝑑𝑥𝑥
Newton’s Law
y
F
b
Fixed plate
13
Definition of Fluid
“Anything that
flows”
It is a substance that deforms continuously when acted on by shearing stress of any magnitude.
Fluids may include:
GASES
LIQUIDS
MOLTEN SOLIDS
17
Definition of Fluid Mechanics
MECHANICS is the branch of science that deals with motion and forces producing motion
MECHANICS
STATICS
DYNAMICS
21
Definition of Fluid Mechanics
MECHANICS is the branch of science that deals with motion and forces producing motion
MECHANICS
𝑭𝑭 = 𝟎𝟎
𝑭𝑭 = 𝒎𝒎𝒂𝒂
23
Definition of Fluid Mechanics
MECHANICS is the branch of science that deals with motion and forces producing motion
MECHANICS
STATICS
DYNAMICS
KINEMATICS
KINEMATIC HOW?
EQUATIONS
KINETICS
WHY?
SUMMATION
OF FORCES
27
Definition of Fluid Mechanics
MECHANICS is the branch of science that deals with motion and forces producing motion
FLUID
MECHANICS
FLUID
STATICS
FLUID
DYNAMICS
KINEMATICS
KINEMATIC HOW?
EQUATIONS
KINETICS
WHY?
SUMMATION
OF FORCES
28
Definition of Fluid Mechanics
MECHANICS is the branch of science that deals with motion and forces producing motion
FLUID
MECHANICS
30
Definition of Fluid Mechanics
31
Newtonian vs Non-Newtonian
Fluids for which the shear stress is linearly related to the rate of shear strain (also referred to
as the rate of angular deformation) are referred to as Newtonian fluids. Otherwise, the fluids
are Non-Newtonian fluids.
τ
U
y
F
b
Fixed plate
du
dy
36
Newtonian vs Non-Newtonian
Fluids for which the shear stress is linearly related to the rate of shear strain (also referred to
as the rate of angular deformation) are referred to as Newtonian fluids. Otherwise, the fluids
are Non-Newtonian fluids.
𝑑𝑑(𝑣𝑣)
𝝉𝝉 = −μ
𝑑𝑑𝑥𝑥
τ
The slope of the shear stress vs. rate of strain is denoted as the
apparent viscosity, μapp.
du
dy
37
Newtonian vs Non-Newtonian
Fluids for which the shear stress is linearly related to the rate of shear strain (also referred to
as the rate of angular deformation) are referred to as Newtonian fluids. Otherwise, the fluids
are Non-Newtonian fluids.
τ
Shear thinning fluids (or “pseudoplastic”) the apparent
viscosity decreases with increasing shear rate – the harder the
fluid is sheared, the less viscous it becomes.
Thixotropic Fluid – time-dependent pseudoplastic
du
dy
40
Newtonian vs Non-Newtonian
Fluids for which the shear stress is linearly related to the rate of shear strain (also referred to
as the rate of angular deformation) are referred to as Newtonian fluids. Otherwise, the fluids
are Non-Newtonian fluids.
τ
Shear thickening fluids (or “dilatant”), the apparent viscosity
increases with increasing shear rate – the harder the fluid is
sheared, the more viscous it becomes.
Rheopectic Fluid – time-dependent dilatant
du
dy
43
Newtonian vs Non-Newtonian
Fluids for which the shear stress is linearly related to the rate of shear strain (also referred to
as the rate of angular deformation) are referred to as Newtonian fluids. Otherwise, the fluids
are Non-Newtonian fluids.
τ
Bingham plastic, is neither a fluid nor a solid. Such material
can withstand a finite shear stress without motion (therefore,
not a fluid), but once the yield point is exceeded, it flows like a
fluid (hence, it is not solid).
du
dy
45
FLUID MECHANICS
Lecture 1
Introduction
and Basic Concepts
Introduction
Statics
of Rigid-bodies
Engineering
Mechanics
Dynamics
of Rigid Bodies
• Kinematics
• Kinetics
Mechanics
of Deformable Bodies
• Fluid Statics
Fluid Dynamics
Fluid Mechanics •
Introduction
Fluid Mechanics
is defined as
the science that deals with
the behavior of fluids at rest (fluid statics)
or in motion (fluid dynamics),
and the interaction of fluids with
solids or other fluids at the boundaries.
Introduction
Fluid Mechanics Categories
Hydrodynamics
The study of the motion of fluids that can be approximated as incompressible (such as
liquids, especially water, and gases at low speeds).
• Hydraulics - a subcategory of Hydrodynamics which deals with liquid flows in pipes
and open channels
Ship Propeller
Introduction
Fluid Mechanics Categories
Gas dynamics
It deals with the flow of fluids that undergo significant density changes, such as the
flow of gases through nozzles at high speeds.
Pulsed Protostellar Jets
Supersonic Flight
Introduction
Fluid Mechanics Categories
Aerodynamics
It deals with the flow of gases (especially air) over bodies such as aircraft, rockets, and
automobiles at high or low speeds.
CFD Model of Airflow on Wing of a Plane
Why are golf balls made with dimpled surfaces?
A Sports Car in a Wind Tunnel
Introduction
Fluid Mechanics Categories
Other specialized categories:
• Meteorology
A branch of the atmospheric sciences which includes atmospheric chemistry and
atmospheric physics, with a major focus on weather forecasting.
Introduction
Fluid Mechanics Categories
Other specialized categories:
• Oceanography
The study of the physical and biological aspects of the ocean.
Introduction
Fluid Mechanics Categories
Other specialized categories:
• Hydrology
The branch of science concerned with the properties of the earth's water, especially its
movement in relation to land.
Introduction
What Is a Fluid?
Fluid
A substance in the liquid or gas phase.
Distinction between a solid and a fluid:
Solids:
• A solid can resist an applied
shear stress by deforming.
• Stress is proportional to strain.
Fluids:
• A fluid deforms continuously
under
the influence of a shear stress,
no matter how small.
• Stress is proportional to strain
rate.
Introduction
What Is a Fluid?
Pressure
The normal stress in a fluid at rest.
Introduction
What Is a Fluid?
Liquid:
• Groups of molecules can move
relative to each other, but the
volume remains relatively
constant because of the strong
cohesive forces between the
molecules.
Gas:
• Gas expands until it encounters
the walls of the container and
fills the entire available space.
This is because the gas
molecules are widely spaced,
and the cohesive forces
between them are very small.
Introduction
What Is a Fluid?
The arrangement of atoms in different phases:
(a) molecules are at relatively fixed positions in a solid,
(b) groups of molecules move about each other in the liquid phase, and
(c) individual molecules move about at random in the gas phase.
The No-Slip Condition
A fluid in direct contact with a solid “sticks” to the surface, and there is no slip.
• Viscosity is the fluid property responsible for the no-slip condition and the
development of the boundary layer.
Classification of Fluid Flows
Viscous versus Inviscid Regions of Flow
Viscous Flows
• Flows in which the frictional effects are significant.
Inviscid Flows
• Flows in which the viscous forces are negligibly small compared
to inertial or pressure forces.
Classification of Fluid Flows
Internal versus External Flow
External Flow:
• flow of an unbounded fluid over
a surface such as a plate, a wire,
or a pipe
Internal Flow:
• flow in a pipe or duct if the fluid
is completely bounded by solid
surfaces.
Classification of Fluid Flows
Compressible versus Incompressible Flow
Incompressible Flow:
• If the density remains nearly
constant throughout
• Liquids are usually referred to as
incompressible substances
Compressible Flow:
• Gases are highly compressible
Classification of Fluid Flows
Laminar versus Turbulent Flow
Laminar Flow:
• The highly ordered fluid motion
characterized by smooth layers
of fluid.
Transitional Flow:
• A flow that alternates between
being laminar and turbulent
Turbulent Flow:
• The highly disordered fluid
motion that typically occurs at
high velocities and is
characterized by velocity
fluctuations.
Classification of Fluid Flows
Natural (or Unforced) versus Forced Flow
Natural Flow:
• Fluid motion is due to natural means such as the buoyancy effect, which
manifests itself as the rise of warmer (and thus lighter) fluid and the fall of
cooler (and thus denser) fluid.
Forced Flow:
• a fluid is forced to flow over a surface or in a pipe by external means such
as a pump or a fan.
Classification of Fluid Flows
Steady versus Unsteady Flow
Steady Flow:
• It implies no change of properties, velocity, temperature, etc., at a point
with time.
• Uniform Flow:
• It implies no change with location over a specified region.
Unsteady Flow:
• The most general term that applies to any flow that is not steady.
• Transient Flow:
• Typically used for developing flows.
• Periodic Flow:
• The kind of unsteady flow in which the flow oscillates about a steady
mean.
Classification of Fluid Flows
One-, Two-, and Three-Dimensional Flows
A flow field is best characterized by its velocity distribution, and thus a flow is
said to be one-, two-, or three-dimensional if the flow velocity varies in one,
two, or three primary dimensions, respectively.
• the variation of velocity in certain directions can be small relative to the
variation in other directions and can be ignored with negligible error.
Reference:
Yunus A. Çengel, John M. Cimbala.
Fluid Mechanics: Fundamentals and Applications - 3rd Edition.
1221 Avenue of the Americas, New York, NY 10020: McGraw-Hill, 2014.
FLUID MECHANICS
Lecture 2
Properties of Fluids
Introduction
Intensive properties:
Properties that are independent of the
mass of the system.
Examples:
• Temperature
• Pressure
• Density
Extensive properties:
Properties whose values depend on the
size—or extent—of the system.
Examples:
• Total Mass
• Total Volume
• Total Momentum
Introduction
Continuum
Continuum:
An idealization, disregarding the
atomic nature of the fluid.
• Continuous – i.e. assumes that
the properties vary continually in
space with no jump
discontinuities
• Homogeneous
• allows properties to be treated
as point functions
A fluid is composed of molecules which
may be widely spaced apart, especially
in the gas phase.
Density and Specific Gravity
Density:
• mass per unit volume.
𝑚𝑚
𝜌𝜌 =
𝑉𝑉
Specific Gravity, or Relative
Density:
• the ratio of the density of a substance to
the density of some standard substance
at a specified temperature (usually water
at 4°C, for which 𝜌𝜌𝐻𝐻2𝑂𝑂 = 1000 kg/m3 ).
𝜌𝜌
𝑆𝑆𝑆𝑆 =
𝜌𝜌𝐻𝐻2𝑂𝑂
Specific Volume:
• volume per unit mass.
𝑉𝑉
𝑣𝑣 =
𝑚𝑚
Specific Weight or Weight Density:
• weight per unit volume.
Density and Specific Gravity
Density of Ideal Gases
The state postulate:
The state of a simple compressible system is completely specified by two
independent, intensive properties.
Ideal-gas equation of state:
𝑃𝑃𝑃𝑃 = 𝑅𝑅𝑅𝑅
𝑃𝑃 = 𝜌𝜌𝑅𝑅𝑅𝑅
Thermodynamic Temperature
Scale:
• SI:
Kelvin (Absolute),
Celsius (Relative)
𝑇𝑇 𝐾𝐾 = 𝑇𝑇 °𝐶𝐶 + 273.15 = 𝑇𝑇 𝑅𝑅 /1.8
• English:
Rankine (Absolute),
Celsius (Relative)
𝑇𝑇 𝑅𝑅 = 𝑇𝑇 °𝐹𝐹 + 459.67 = 1.8 𝑇𝑇 𝐾𝐾
Universal Gas Constant:
J
𝑅𝑅𝑢𝑢 = 8.31447
mol · K
BTU
𝑅𝑅𝑢𝑢 = 1.986
lb · mol · R
Vapor Pressure and Cavitation
Saturation Temperature, 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 :
• The temperature at which a pure
substance changes phase at a given
pressure.
Saturation Pressure, 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 :
• The pressure at which a pure substance
changes phase at a given temperature.
Vapor pressure 𝑃𝑃𝑣𝑣
of a pure substance:
• The pressure exerted by its vapor in phase
equilibrium with its liquid at a given
temperature.
Partial pressure
• The pressure of a gas or vapor in a
mixture with other gases.
Vapor Pressure and Cavitation
Cavitation
• The liquid pressure in liquid-flow systems dropping below the vapor
pressure at some locations, resulting in unplanned vaporization.
• The vapor bubbles (called cavitation bubbles since they form
“cavities” in the liquid) collapse as they are swept away from the lowpressure regions, generating highly destructive, extremely highpressure waves.
Energy and Specific Heats
Total energy, 𝐸𝐸 (or 𝑒𝑒 on a unit mass basis)
• The sum of energy that can exist in numerous forms such as thermal,
mechanical, kinetic, potential, electrical, magnetic, chemical, and
nuclear.
Internal Energy,
Kinetic Energy,
𝑈𝑈 (or 𝑢𝑢 on a unit mass basis)
𝐾𝐾𝐾𝐾 (or 𝑘𝑘𝑘𝑘 on a unit mass basis)
• The sum of all microscopic
• The energy that a system
forms of energy
possesses as a result of its
motion.
1 2
Enthalpy,
𝑘𝑘𝑘𝑘 = 𝑣𝑣
𝐻𝐻 (or ℎ on a unit mass basis)
2
• The combination of properties 𝑢𝑢
Potential Energy,
(Internal Energy) and 𝑃𝑃𝑃𝑃 (flow
𝑃𝑃𝑃𝑃 (or 𝑝𝑝𝑝𝑝 on a unit mass basis)
energy or flow work)
• The energy that a system
ℎ = 𝑢𝑢 + 𝑃𝑃𝑃𝑃
possesses as a result of its
𝑃𝑃
ℎ = 𝑢𝑢 +
elevation in a gravitational field
𝜌𝜌
𝑝𝑝𝑝𝑝 = 𝑔𝑔𝑔𝑔
Energy and Specific Heats
Internal energy, 𝑢𝑢
• represents the microscopic energy
of a nonflowing fluid per unit mass.
• The differential and finite changes in
the internal energy:
𝑑𝑑𝑑𝑑 = 𝑐𝑐𝑣𝑣 𝑑𝑑𝑑𝑑
Enthalpy, ℎ
• represents the microscopic energy
of a flowing fluid per unit mass.
• The differential and finite changes in
the enthalpy:
𝑑𝑑ℎ = 𝑐𝑐𝑝𝑝 𝑑𝑑𝑑𝑑
• For incompressible substances,
the constant-volume and constant-pressure specific heats are identical
𝑐𝑐𝑝𝑝 ≅ 𝑐𝑐𝑣𝑣 ≅ 𝑐𝑐
Compressibility and Speed of Sound
Coefficient of Compressibility
Coefficient of Compressibility, 𝜅𝜅
(also called the Bulk Modulus of Compressibility
or Bulk Modulus of Elasticity):
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
= 𝜌𝜌
Pa
𝜅𝜅 = −𝑣𝑣
𝜕𝜕𝜕𝜕 𝑇𝑇
𝜕𝜕𝜕𝜕 𝑇𝑇
• in terms of finite changes:
∆𝑃𝑃
∆𝑃𝑃
≅
𝜅𝜅 ≅ −
𝑇𝑇 = c𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
∆𝑣𝑣/𝑣𝑣 ∆𝜌𝜌/𝜌𝜌
• a large value of 𝜅𝜅 indicates that a large change in
pressure is needed to cause a small fractional
change in volume
• a fluid with a large 𝜅𝜅 is essentially incompressible
Compressibility and Speed of Sound
Coefficient of Compressibility
For an ideal gas:
• thus:
𝑃𝑃 = 𝜌𝜌𝑅𝑅𝑅𝑅
𝜕𝜕𝜕𝜕
𝑃𝑃
= 𝑅𝑅𝑅𝑅 =
𝜕𝜕𝜕𝜕 𝑇𝑇
𝜌𝜌
𝜅𝜅ideal gas = 𝑃𝑃
Pa
• the coefficient of compressibility of an ideal gas
is equal to its absolute pressure
• the coefficient of compressibility of the gas
increases with increasing pressure
Compressibility and Speed of Sound
Coefficient of Compressibility
Isothermal Compressibility, 𝛼𝛼:
1
1 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
𝛼𝛼 = = −
=
𝑣𝑣 𝜕𝜕𝜕𝜕 𝑇𝑇 𝜌𝜌 𝜕𝜕𝜕𝜕 𝑇𝑇
𝜅𝜅
1/Pa
• it represents the fractional change in volume or
density corresponding to a unit change in
pressure
Compressibility and Speed of Sound
Coefficient of Volume Expansion
Coefficient of Volume Expansion, 𝛽𝛽
(or Volume Expansivity):
1 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
=−
1/K
𝛽𝛽 =
𝑣𝑣 𝜕𝜕𝑇𝑇 𝑃𝑃
𝜌𝜌 𝜕𝜕𝑇𝑇 𝑃𝑃
• in terms of finite changes:
∆𝑣𝑣/𝑣𝑣
∆𝜌𝜌/𝜌𝜌
≅−
𝛽𝛽 ≅
𝑃𝑃 = c𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
∆𝑇𝑇
∆𝑇𝑇
• a large value of 𝛽𝛽 for a fluid means a large change
in density with temperature
• the product 𝛽𝛽 ∆𝑇𝑇 represents the fraction of
volume change of a fluid that corresponds to a
temperature change of ∆𝑇𝑇 at constant pressure
Compressibility and Speed of Sound
Coefficient of Volume Expansion
For an ideal gas:
• thus:
𝑃𝑃𝑣𝑣 = 𝑅𝑅𝑅𝑅
𝜕𝜕𝜕𝜕
𝑅𝑅 𝑣𝑣
= =
𝜕𝜕𝜕𝜕 𝑃𝑃 𝑃𝑃 𝑇𝑇
𝛽𝛽ideal gas =
1
𝑇𝑇
1/K
• the volume expansion coefficient of an ideal gas
at a temperature 𝑇𝑇 is equivalent to the inverse of
the temperature
Compressibility and Speed of Sound
The combined effects of pressure and temperature changes on the
volume change of a fluid can be determined by taking the specific volume
to be a function of T and P.
𝜕𝜕𝜕𝜕
𝜕𝜕𝑣𝑣
𝑑𝑑𝑑𝑑 =
𝑑𝑑𝑑𝑑 +
𝑑𝑑𝑃𝑃 = 𝛽𝛽 𝑑𝑑𝑑𝑑 − 𝛼𝛼 𝑑𝑑𝑑𝑑 𝑣𝑣
𝜕𝜕𝑇𝑇 𝑃𝑃
𝜕𝜕𝑃𝑃 𝑇𝑇
The fractional change in volume (or density) due to changes in pressure
and temperature:
∆𝑣𝑣
∆𝜌𝜌
=−
≅ 𝛽𝛽 ∆𝑇𝑇 − 𝛼𝛼 ∆𝑃𝑃
𝑣𝑣
𝜌𝜌
Compressibility and Speed of Sound
Speed of Sound and Mach Number
Speed of Sound (or Sonic Speed), 𝑐𝑐:
𝑐𝑐 = 𝑘𝑘𝑘𝑘𝑘𝑘
• the speed of sound in a specified
ideal gas is a function of temperature
alone
Compressibility and Speed of Sound
Speed of Sound and Mach Number
Mach Number, Ma:
𝑉𝑉
Ma =
𝑐𝑐
• named after the Austrian
physicist Ernst Mach (1838–
1916)
• the ratio of the actual speed of
the fluid (or an object in still
fluid) to the speed of sound in
the same fluid at the same state
Fluid flow regimes are often described in terms of the flow Mach number:
• Sonic (Ma = 1)
• Subsonic (Ma < 1)
• Supersonic (Ma > 1)
• Hypersonic (Ma ≫ 1)
• Transonic (Ma ≅ 1)
Viscosity
Viscosity:
• a property that represents the
internal resistance of a fluid to
motion
• The viscosity of liquids
decreases with temperature,
whereas the viscosity of gases
increases with temperature
Dynamic Viscosity
(or Coefficient of Viscosity), 𝜇𝜇:
• Units:
N·s
kg
• Pa · s = 2 =
m
m·s
• Poise = 0.1 Pa · s
• Centipoise = 0.01 Poise
Kinematic Viscosity, 𝜈𝜈:
• the ratio of dynamic viscosity to
density
𝜇𝜇
𝜈𝜈 =
𝜌𝜌
• Units:
•
•
m2
s
cm2
m2
Stoke = 1
= 0.0001
s
s
Viscosity
Newtonian Fluids:
• Named after Sir Isaac Newton, who
expressed it first in 1687
• Fluids for which the rate of
deformation is linearly proportional
to the shear stress
• Shear Stress:
𝑑𝑑𝑑𝑑
𝜏𝜏 = 𝜇𝜇
𝑑𝑑𝑑𝑑
• Examples:
• Water
• Air
• Gasoline
• Oils
N/m2
Viscosity
Non-Newtonian Fluids:
• the relationship between shear
stress and rate of deformation is not
linear
• The slope of the curve 𝜏𝜏 versus
𝑑𝑑𝑑𝑑
is
𝑑𝑑𝑑𝑑
referred to as the apparent viscosity
Dilatants
(or Shear Thickening Fluids):
• Fluids for which the apparent
viscosity increases with the rate of
deformation
• Examples:
• Suspended Starch
• Sand
Viscosity
Non-Newtonian Fluids:
• the relationship between shear
stress and rate of deformation is not
linear
• The slope of the curve 𝜏𝜏 versus
𝑑𝑑𝑑𝑑
is
𝑑𝑑𝑑𝑑
referred to as the apparent viscosity
Pseudoplastics
(or Shear Thinning Fluids):
• Fluid which becomes less viscous as
it is sheared harder
• Examples:
• Some Paints
• Polymer Solutions
• Fluids with Suspended Particles
Viscosity
Non-Newtonian Fluids:
• the relationship between shear
stress and rate of deformation is not
linear
• The slope of the curve 𝜏𝜏 versus
𝑑𝑑𝑑𝑑
is
𝑑𝑑𝑑𝑑
referred to as the apparent viscosity
Bingham Plastics
after Eugene C. Bingham (1878–1945):
• Fluids that can resist a finite shear
stress and thus behave as a solid,
but deform continuously when the
shear stress exceeds the yield stress
and behave as a fluid.
• Example:
• Toothpaste
Viscosity
Liquids:
• The viscosity of liquids decreases
with temperature
• The molecules possess more energy
at higher temperatures, and they can
oppose the large cohesive
intermolecular forces more strongly.
Gases:
• The viscosity of gases increases with
temperature
• The kinetic theory of gases predicts
the viscosity of gases to be
proportional to the square root of
temperature:
𝜇𝜇𝑔𝑔𝑔𝑔𝑔𝑔 ∝ 𝑇𝑇
Viscosity
Viscosity
Torque on a Fluid with Concentric
Cylinders, 𝑇𝑇:
2𝜋𝜋𝑅𝑅3 𝜔𝜔𝜔𝜔
4𝜋𝜋 2 𝑅𝑅3 𝑛𝑛𝐿𝐿
̇
= 𝜇𝜇
𝑇𝑇 = 𝐹𝐹𝐹𝐹 = 𝜇𝜇
𝑙𝑙
𝑙𝑙
•
•
•
•
•
𝑅𝑅 = radius of the inner circle
𝜔𝜔 = angular velocity
𝐿𝐿 = Length of the cylinder
𝑙𝑙 = layer of thickness of fluid
𝑛𝑛̇ = number of revolutions per unit
time
Surface Tension and Capillary Effect
Surface Tension (or
Coefficient of Surface
Tension), 𝜎𝜎𝑠𝑠 :
• The pulling effect on the
liquid molecules at an
interface caused by the
attractive forces of
molecules per unit
length
• The stretching work that
needs to be done to
increase the surface
area of the liquid by a
unit amount
• Units:
N·m
J
N
•
=
=
2
2
m
m
m
Surface Tension and Capillary Effect
Surface tension determines the size
of the liquid droplets that form.
Droplet or Air Bubble:
2𝜋𝜋𝑅𝑅 𝜎𝜎𝑠𝑠 = 𝜋𝜋𝑅𝑅2 ∆𝑃𝑃droplet
2𝜎𝜎𝑠𝑠
∆𝑃𝑃droplet = 𝑃𝑃𝑖𝑖 − 𝑃𝑃𝑜𝑜 =
𝑅𝑅
Soap Bubble:
2 2𝜋𝜋𝑅𝑅 𝜎𝜎𝑠𝑠 = 𝜋𝜋𝑅𝑅2 ∆𝑃𝑃droplet
4𝜎𝜎𝑠𝑠
∆𝑃𝑃bubble = 𝑃𝑃𝑖𝑖 − 𝑃𝑃𝑜𝑜 =
𝑅𝑅
Surface Tension and Capillary Effect
Capillary Effect
Capillary Effect:
• a consequence of surface tension,
• the rise or fall of a liquid in a smalldiameter tube inserted into the
liquid.
Contact (or Wetting) Angle, 𝜙𝜙:
• the angle that the tangent to the
liquid surface makes with the solid
surface at the point of contact
• a liquid is said to wet the surface
when 𝜙𝜙 < 90° and not to wet the
surface when 𝜙𝜙 > 90°
Meniscus:
• The curved free surface of a liquid in
a capillary tube
Surface Tension and Capillary Effect
Capillary Effect
Capillary Rise:
2𝜎𝜎𝑠𝑠
cos 𝜙𝜙
ℎ=
𝜌𝜌𝜌𝜌𝜌𝜌
• the thinner the tube is, the greater the rise (or
fall) of the liquid in the tube.
Formulas
• 𝑆𝑆𝑆𝑆 =
𝜌𝜌
𝜌𝜌𝐻𝐻2𝑂𝑂
• 𝑃𝑃 = 𝜌𝜌𝑅𝑅𝑅𝑅
𝜕𝜕𝜕𝜕
= 𝜌𝜌
𝜕𝜕𝜕𝜕 𝑇𝑇
1 𝜕𝜕𝜕𝜕
1
𝛽𝛽 =
=−
𝑣𝑣 𝜕𝜕𝜕𝜕 𝑃𝑃
𝜌𝜌
• 𝜅𝜅 = −𝑣𝑣
•
• 𝑐𝑐 = 𝑘𝑘𝑘𝑘𝑘𝑘
𝑉𝑉
• Ma =
•
𝑐𝑐
𝑑𝑑𝑑𝑑
𝜏𝜏 = 𝜇𝜇
𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕
∆𝑃𝑃
∆𝑃𝑃
≅−
≅
𝜕𝜕𝜕𝜕 𝑇𝑇
∆𝑣𝑣/𝑣𝑣
∆𝜌𝜌/𝜌𝜌
𝜕𝜕𝜕𝜕
∆𝑣𝑣/𝑣𝑣
∆𝜌𝜌/𝜌𝜌
≅
≅−
𝜕𝜕𝜕𝜕 𝑃𝑃
∆𝑇𝑇
∆𝑇𝑇
2𝜎𝜎
• ∆𝑃𝑃droplet = 𝑃𝑃𝑖𝑖 − 𝑃𝑃𝑜𝑜 = 𝑅𝑅 𝑠𝑠
• ℎ=
2𝜎𝜎𝑠𝑠
cos 𝜙𝜙
𝜌𝜌𝜌𝜌𝜌𝜌
∆𝑃𝑃bubble = 𝑃𝑃𝑖𝑖 − 𝑃𝑃𝑜𝑜 =
4𝜎𝜎𝑠𝑠
𝑅𝑅
References
Yunus A. Çengel, John M. Cimbala
Fluid Mechanics: Fundamentals and Applications - 3rd Edition.
1221 Avenue of the Americas, New York, NY 10020: McGraw-Hill, 2014.
FLUID MECHANICS
Properties of Fluids
Lecture 2
Problems
Density, Specific Gravity, and Mass of Air in a Room
Determine the density, specific gravity, and mass of the air in a room whose
dimensions are 4 m x 5 m x 6 m at 100 kPa and 25°C.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨:
At specified conditions, air can be treated as an ideal gas.
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
The gas constant of air is:
N·m
𝑅𝑅 = 287.08
kg · K
𝑼𝑼𝑼𝑼𝑼𝑼𝑼𝑼𝑼𝑼 𝒕𝒕𝒕𝒕𝒕𝒕 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝒐𝒐𝒐𝒐 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺:
𝑚𝑚
𝑃𝑃
=
𝑉𝑉 𝑅𝑅𝑅𝑅
The density of the air is:
𝑃𝑃𝑃𝑃 = 𝑚𝑚𝑚𝑚𝑚𝑚
→
→
𝜌𝜌 =
𝑃𝑃
𝑅𝑅𝑅𝑅
100,000
𝑃𝑃
100 kPa
=
𝜌𝜌 =
=
N·m
N·m
𝑅𝑅𝑅𝑅
287.08
25 + 273.15 K
287.08
kg · K
kg · K
𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒇𝒇𝒇𝒇𝒇𝒇 𝒕𝒕𝒕𝒕𝒕𝒕 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈𝒈 𝒐𝒐𝒐𝒐 𝒕𝒕𝒕𝒕𝒕𝒕 𝑨𝑨𝑨𝑨𝑨𝑨
SG =
𝜌𝜌
𝜌𝜌H2O
=
1.168321177
1000
kg
m3
kg
m3 = 0.001168321177
N
m2
kg
= 1.168321177 3
m
298.15 K
𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒇𝒇𝒇𝒇𝒇𝒇 𝒕𝒕𝒕𝒕𝒕𝒕 𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝒐𝒐𝒐𝒐 𝒕𝒕𝒕𝒕𝒕𝒕 𝑨𝑨𝑨𝑨𝑨𝑨
𝜌𝜌 =
Answer:
𝜌𝜌 = 1.17
𝑚𝑚
𝑉𝑉
→
𝑚𝑚 = 𝜌𝜌𝑉𝑉
→
𝑚𝑚 = 1.168321177
𝑚𝑚 = 140.1985412 kg
kg
, SG = 0.00117, 𝑚𝑚 = 140 kg
m3
Example 2-1 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
kg
m3
4m 5m 6m
Compressibility
Water at 15°C and 1 atm pressure is heated to 95°C at constant pressure. Using coefficient of volume expansion data,
determine the change in the density of water. (The density of water at 15°C and 1 atm pressure is 𝜌𝜌 = 999.1 kg/m3 .
The coefficient of volume expansion at the average temperature of (15 + 95)/2 = 55°C is 𝛽𝛽 = 0.484 10−3 K −1 ).
𝑻𝑻𝑻𝑻𝑻𝑻 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒐𝒐𝒐𝒐 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬
The Volume Expansivity is a function of Temperature and Specific Volume or Density:
∆𝑣𝑣/𝑣𝑣
∆𝜌𝜌/𝜌𝜌
𝑜𝑜𝑜𝑜 𝛽𝛽 = −
𝛽𝛽 =
∆𝑇𝑇
∆𝑇𝑇
Since the data given is the Density of water, use the 2nd form, and rearrange:
∆𝜌𝜌 = −𝜌𝜌𝛽𝛽∆𝑇𝑇
Plugging the values:
∆𝜌𝜌 = − 999.1 kg/m3 0.484 10−3 K −1 95°C − 15°C
∆𝜌𝜌 = −38.685152 kg/m3
Answer:
kg
∆𝜌𝜌 = −38.7 3
m
Problem 2-40 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Compressibility
It is observed that the density of an ideal gas increases by 10 percent when compressed isothermally
from 10 atm to 11 atm. Determine the percent increase in density of the gas if it is compressed isothermally
from 1000 atm to 1001 atm.
𝑻𝑻𝑻𝑻𝑻𝑻 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒐𝒐𝒐𝒐 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪:
The Coefficient of Compressibility or Bulk Modulus of Elasticity is a function of Pressure and Specific Volume or Density
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝜅𝜅 = −𝑣𝑣
𝑜𝑜𝑜𝑜 𝜅𝜅 = 𝜌𝜌
𝜕𝜕𝜕𝜕 𝑇𝑇
𝜕𝜕𝜕𝜕 𝑇𝑇
Since the Density of an ideal gas is to be determined, use the 2nd form, and substitute the equation of state:
𝜅𝜅 = 𝜌𝜌 𝑅𝑅𝑅𝑅 = 𝑃𝑃
When the pressure is 10 atm:
𝜅𝜅 = 10 atm
When the pressure is 1000 atm:
1001 atm − 1000 atm
∆𝜌𝜌
1 atm
𝜅𝜅 = 1000 atm =
→
=
= 0.001
∆𝜌𝜌/𝜌𝜌
𝜌𝜌
1000 atm
Answer:
At 10 atm: 10%
At 1000 atm: 0.1%
Problem 2-41 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Compressibility
Using the definition of the coefficient of volume expansion and the expression 𝛽𝛽ideal gas = 1/𝑇𝑇, show that the
percent increase in the specific volume of an ideal gas during isobaric expansion is equal to the percent increase in
absolute temperature.
𝑻𝑻𝑻𝑻𝑻𝑻 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒐𝒐𝒐𝒐 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬
The Volume Expansivity is a function of Temperature and Specific Volume or Density:
1 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
𝛽𝛽 =
𝑜𝑜𝑜𝑜 𝛽𝛽 = −
𝑣𝑣 𝜕𝜕𝜕𝜕 𝑃𝑃
𝜌𝜌 𝜕𝜕𝜕𝜕 𝑃𝑃
Since the Density of an ideal gas is to be determined, use the 1st form, and substitute 𝛽𝛽ideal gas = 1/𝑇𝑇:
1 1 𝜕𝜕𝜕𝜕
=
𝑇𝑇 𝑣𝑣 𝜕𝜕𝜕𝜕 𝑃𝑃
In terms of finite changes:
1 1 ∆𝑣𝑣
=
𝑇𝑇 𝑣𝑣 ∆𝑇𝑇
Answer:
∆𝑣𝑣 ∆𝑇𝑇
=
𝑣𝑣
𝑇𝑇
Problem 2-42 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Compressibility
Water at 1 atm pressure is compressed to 400 atm pressure isothermally.
Determine the increase in the density of water. Take the isothermal compressibility of water to be
4.80 10−5 atm−1 . (The density of water at 20°C and 1 atm pressure is 𝜌𝜌 = 998 kg/m3 )
𝑻𝑻𝑻𝑻𝑻𝑻 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪
The Isothermal Compressibility is a function of Pressure and Specific Volume or Density:
1 𝜕𝜕𝜕𝜕
1 𝜕𝜕𝜕𝜕
𝛼𝛼 = −
𝑜𝑜𝑜𝑜 𝛼𝛼 =
𝑣𝑣 𝜕𝜕𝜕𝜕 𝑇𝑇
𝜌𝜌 𝜕𝜕𝜕𝜕 𝑇𝑇
Since the Density of the water is to be determined, use the 2nd form, and use finite changes:
1 ∆𝜌𝜌
𝛼𝛼 =
→ ∆𝜌𝜌 = 𝛼𝛼𝜌𝜌 ∆𝑃𝑃
𝜌𝜌 ∆𝑃𝑃
Plugging the values:
∆𝜌𝜌 = 4.80 10−5 atm−1 998 kg/m3 400 atm − 1 atm
∆𝜌𝜌 = 19.113696 kg/m3
Answer:
∆𝜌𝜌 = 19.2 kg/m3
Problem 2-43 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Variation of Density with Temperature and Pressure
Consider water initially at 20°C and 1 atm. Determine the final density of the water (a) if it is heated to 50°C at a
constant pressure of 1 atm, and (b) if it is compressed to 100-atm pressure at a constant temperature of 20°C.
Take the isothermal compressibility of water to be 𝛼𝛼 = 4.80 10−5 atm−1 . 𝜌𝜌𝐻𝐻2 𝑂𝑂 = 998 kg/m3 @ 20°C.
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
At the average given temperature (50°𝐶𝐶 + 25 °𝐶𝐶)/2 = 35 °𝐶𝐶, the coefficient of volume expansion:
𝛽𝛽 ≈ 0.000337/𝐾𝐾
𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒊𝒊𝒊𝒊 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽
∆𝜌𝜌
∆𝜌𝜌
−
≅ 𝛽𝛽 ∆𝑇𝑇 − 𝛼𝛼 ∆𝑃𝑃 →
= 𝛼𝛼 ∆𝑃𝑃 − 𝛽𝛽 ∆𝑇𝑇
𝜌𝜌
𝜌𝜌
∆𝜌𝜌 = 𝜌𝜌 𝛼𝛼 ∆𝑃𝑃 − 𝛽𝛽 ∆𝑇𝑇
𝒂𝒂 𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮 𝒂𝒂 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 @ 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑,
∆𝜌𝜌 = 𝜌𝜌 𝛼𝛼 ∆𝑃𝑃 − 𝛽𝛽 ∆𝑇𝑇
∆𝜌𝜌 = 998 kg/m3 4.80 10−5 atm−1 0 − 0.000337/𝐾𝐾 50 K − 20 K
∆𝜌𝜌 = 𝜌𝜌𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝜌𝜌𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = −10.0 kg/m3
𝜌𝜌𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 998 kg/m3 + −10.0 kg/m3
𝒃𝒃 𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮𝑮 𝒂𝒂 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 @ 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕,
∆𝜌𝜌 = 998 kg/m3
∆𝜌𝜌 = 𝜌𝜌 𝛼𝛼 ∆𝑃𝑃 − 𝛽𝛽 ∆𝑇𝑇
4.80 10−5 atm−1 100 atm − 1 atm − 0.000337/𝐾𝐾 293.15 K − 293.15 K
∆𝜌𝜌 = 𝜌𝜌𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝜌𝜌𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 4.7 kg/m3
𝜌𝜌𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 998 kg/m3 + 4.7 kg/m3
Answer:
𝜌𝜌50°𝐶𝐶,1atm = 988.0 kg/m3 , 𝜌𝜌20°𝐶𝐶,100atm = 1002.7 kg/m3
Example 2-3 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Mach Number of Air Entering a Diffuser
Air enters a diffuser shown with a speed of 200 m/s. Determine (a) the speed of sound and (b) the Mach number at
the diffuser inlet when the air temperature is 30°C.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
Air at the specified conditions behaves as an ideal gas.
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
Gas constant:
J
kg · K
Specific Heat ratio:
𝑘𝑘 = 1.4
𝑅𝑅 = 287.08
𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒐𝒐𝒐𝒐 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺
𝑐𝑐 = 𝑘𝑘𝑘𝑘𝑘𝑘 =
𝑐𝑐 =
1.4
1.4
kg · m2
2
s
287.08
kg · K
J
287.08
kg · K
273.15 + 30 K
273.15 + 30 K = 349.0553291 m/s
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵
Ma =
Answer:
𝑐𝑐 = 349 m/s, Ma = 0.573
𝑉𝑉
200 m/s
=
𝑐𝑐 349.0553291 m/s
Ma = 0.5729750653
Example 2-4 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Speed of Sound
The Airbus A-340 passenger plane has a maximum takeoff weight of about 260,000 kg, a length of 64 m, a wing span
of 60 m, a maximum cruising speed of 945 km/h, a seating capacity of 271 passengers, a maximum cruising altitude
of 14,000 m, and a maximum range of 12,000 km. The air temperature at the crusing altitude is about
-60°C. Determine the Mach number of this plane for the stated limiting conditions.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
Air at the specified conditions behaves as an ideal gas.
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
Gas constant:
J
kg · K
Specific Heat ratio:
𝑘𝑘 = 1.4
𝑅𝑅 = 287.08
𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒐𝒐𝒐𝒐 𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺
𝑐𝑐 = 𝑘𝑘𝑘𝑘𝑘𝑘 =
𝑐𝑐 =
1.4
1.4
kg · m2
s2
287.08
kg · K
J
287.08
kg · K
273.15 − 60 K
273.15 − 60 K = 292.6901823 m/s
Answer:
Ma = 0.897
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵
Ma =
𝑉𝑉
=
𝑐𝑐
945
km 1 h
1000 m
h 3600 s
1 km
292.6901823 m/s
Ma = 0.8968527675
Problem 2-62 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Viscosity
A 50-cm x 30-cm x 20-cm block weighing 150 N is to be moved at a constant velocity of 0.80 m/s on an inclined
surface with a friction coefficient of 0.27. (a) Determine the force F that needs to be applied in the horizontal
direction. (b) If a 0.40-mm-thick oil film with a dynamic viscosity of 0.012 Pa · s is applied between the block and
inclined surface, determine the percent reduction in the required force.
𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝟏𝟏: 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝒐𝒐𝒐𝒐 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬
x-y axes are rotated 20° counterclockwise:
� 𝐹𝐹𝑥𝑥 = 0
𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 − 𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫:
𝑁𝑁𝐴𝐴
Friction
𝑊𝑊
𝐹𝐹1
𝑉𝑉
→
� 𝐹𝐹𝑦𝑦 = 0 →
𝑬𝑬𝑬𝑬. 𝟏𝟏 = 𝑬𝑬𝑬𝑬. (𝟐𝟐)
𝐹𝐹1 cos 20° − 150 N sin 20°
= 𝐹𝐹1 sin 20° + 150 N cos 20°
0.27
𝐹𝐹1 cos 20° − 0.27 sin 20° = 150 N sin 20° + 0.27 cos 20°
→
𝐹𝐹1 = 150 N
𝐹𝐹1 cos 20° − 𝑊𝑊 sin 20° − Friction = 0
𝐹𝐹1 cos 20° − 150 N sin 20° − 0.27𝑁𝑁 = 0
𝐹𝐹1 cos 20° − 150 N sin 20°
𝐸𝐸𝐸𝐸. 1
𝑁𝑁 =
0.27
𝑁𝑁 − 𝐹𝐹1 sin 20° − 𝑊𝑊 cos 20° = 0
𝑁𝑁 = 𝐹𝐹1 sin 20° + 𝑊𝑊 cos 20°
𝑁𝑁 = 𝐹𝐹1 sin 20° + 150 N cos 20°
sin 20° + 0.27 cos 20°
cos 20° − 0.27 sin 20°
Answer:
𝐹𝐹1 = 105.5 N, 𝐹𝐹2 = 57.2 N, 45.8% Percent Reduction
Problem 2-79 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
→
𝐸𝐸𝐸𝐸. 2
𝐹𝐹1 = 105.4592197 N
Viscosity
A 50-cm x 30-cm x 20-cm block weighing 150 N is to be moved at a constant velocity of 0.80 m/s on an inclined
surface with a friction coefficient of 0.27. (a) Determine the force F that needs to be applied in the horizontal
direction. (b) If a 0.40-mm-thick oil film with a dynamic viscosity of 0.012 Pa · s is applied between the block and
inclined surface, determine the percent reduction in the required force.
𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝟐𝟐: 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭
𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 − 𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫𝑫:
𝑁𝑁𝐴𝐴
Friction
𝑊𝑊
Convert in terms of finite changes:
𝑑𝑑𝑑𝑑
𝜏𝜏 = 𝜇𝜇
𝑑𝑑𝑑𝑑
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
Friction
∆𝑢𝑢
= 𝜇𝜇
𝐴𝐴
∆𝑦𝑦
m
0.8
s
Friction = 0.012 Pa · s 0.5 m 0.2 m
1m
0.40 mm
1000 mm
𝑉𝑉
𝐹𝐹2
𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝒐𝒐𝒐𝒐 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬
x-y axes are rotated 20° counterclockwise:
𝐹𝐹2 cos 20° − 𝑊𝑊 sin 20° − Friction = 0
𝐹𝐹2 =
150 N sin 20° + 2.4 N
cos 20°
→
→
𝐹𝐹2 =
𝑊𝑊 sin 20° + Friction
cos 20°
𝐹𝐹2 = 57.14956179 N
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹
∆𝑢𝑢
∆𝑦𝑦
→
𝜏𝜏 = 𝜇𝜇
→
∆𝑢𝑢
Friction = 𝜇𝜇𝜇𝜇
∆𝑦𝑦
→
Friction = 2.4 N
𝐹𝐹1 − 𝐹𝐹2
%𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 =
𝐹𝐹1
105.4592197 N − 57.14956179 N
= 0.4580885203
%𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 =
105.4592197 N
Answer:
𝐹𝐹1 = 105.5 N, 𝐹𝐹2 = 57.2 N, 45.8% Percent Reduction
Problem 2-79 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Viscosity
A thin 30-cm x 30-cm flat plate is pulled at 3 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between
two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown.
The dynamic viscosity of the oil is 0.027 Pa·s.
Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil
velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion.
𝑻𝑻𝑻𝑻𝑻𝑻 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
𝑦𝑦𝐴𝐴 , 𝑉𝑉0
3 m/s
0.3 m/s
𝒂𝒂 𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳𝑳 𝒐𝒐𝒐𝒐 𝒁𝒁𝒁𝒁𝒁𝒁𝒁𝒁 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽
Use the ratio and proportion between the velocity gradient and vertical displacement:
𝑦𝑦𝐴𝐴
𝑉𝑉 − 𝑉𝑉𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
=
ℎ𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑉𝑉𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 − 𝑉𝑉𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
If the velocity is zero:
𝑦𝑦𝐴𝐴
0 − −0.3 m/s
0.3 m/s
13
=
→ 𝑦𝑦𝐴𝐴 = 2.6 mm
→ 𝑦𝑦𝐴𝐴 =
mm
2.6 mm 3 m/s − −0.3 m/s
3.3 m/s
55
Answer:
𝑦𝑦𝐴𝐴 = 0.236 mm, 𝐹𝐹 = 10.4 N
Problem 2-81 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Viscosity
A thin 30-cm x 30-cm flat plate is pulled at 3 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between
two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown.
The dynamic viscosity of the oil is 0.027 Pa·s.
Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil
velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion.
𝑻𝑻𝑻𝑻𝑻𝑻 𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
𝑦𝑦𝐴𝐴 , 𝑉𝑉0
3 m/s
0.3 m/s
𝒃𝒃 𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 𝒕𝒕𝒕𝒕 𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 𝒕𝒕𝒕𝒕𝒕𝒕 𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴
From the Equation of Equilibrium, the Force needed for constant velocity is equal to the total friction of the fluid:
𝐹𝐹 = 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡 + 𝐹𝐹𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
∆𝑢𝑢𝑡𝑡𝑡𝑡𝑡𝑡
∆𝑢𝑢𝑡𝑡𝑡𝑡𝑡𝑡 ∆𝑢𝑢𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
∆𝑢𝑢𝑏𝑏𝑏𝑏𝑏𝑏𝑡𝑡𝑡𝑡𝑚𝑚
𝐹𝐹 = 𝐴𝐴 𝜏𝜏𝑡𝑡𝑡𝑡𝑡𝑡 + 𝜏𝜏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 𝐴𝐴 𝜇𝜇
+ 𝜇𝜇
= 𝐴𝐴𝐴𝐴
+
∆𝑦𝑦𝑡𝑡𝑡𝑡𝑡𝑡
∆𝑦𝑦𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
ℎ1
ℎ2
3 m/s
3 m/s − −0.3 m/s
𝐹𝐹 = 0.3 m 0.3 m 0.027 Pa · s
+
= 10.37423077 N
0.001 m
0.0026 m
Answer:
𝑦𝑦𝐴𝐴 = 0.236 mm, 𝐹𝐹 = 10.4 N
Problem 2-81 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Determining the Viscosity of a Fluid
The viscosity of a fluid is to be measured by a viscometer constructed of two 40-cm-long concentric cylinders. The
outer diameter of the inner cylinder is 12 cm, and the gap between the two cylinders is 0.15 cm. The inner cylinder is
rotated at 300 rpm, and the torque is measured to be 1.8 N·m. Determine the viscosity of the fluid.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
• The inner cylinder is completely submerged in the fluid.
• The viscous effects on the two ends of the inner cylinder are negligible.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
the profile can be approximated as being linear since:
𝑙𝑙
= 0.025 ≪ 1
𝑅𝑅
Plugging the values:
𝑇𝑇𝑇𝑇
1.8 N · m 0.0015 m
𝜇𝜇 = 2 3 =
1
1 min
̇
4𝜋𝜋 𝑅𝑅 𝑛𝑛𝐿𝐿
4𝜋𝜋 2 0.06 m 3 300 min 60 s
𝜇𝜇 = 0.1583143494 N · s/m2
0.4 m
Answer:
𝜇𝜇 = 0.158 N · s/m2
Example 2-5 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
The Capillary Rise of Water in a Tube
A 0.6-mm-diameter glass tube is inserted into water at 20°C in a cup.
Determine the capillary rise of water in the tube.
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
• There are no impurities in the water and no contamination on the surfaces of the glass tube.
• The experiment is conducted in atmospheric air.
𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷𝑷
The surface tension of water at 20°C:
𝜎𝜎𝑠𝑠 = 0.073 N/m
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨
Solving for the capillary rise:
ℎ=
ℎ=
1000 kg/m3
1000 kg/m3
2𝜎𝜎𝑠𝑠
cos 𝜙𝜙
ℎ=
𝜌𝜌𝜌𝜌𝜌𝜌
2 0.073 N/m
0.6 mm
9.80665 m/s 2
2
kg · m
/m
s2
0.6 mm
9.80665 m/s 2
2
2 0.073
ℎ = 0.04962618903 m
1m
1000 mm
1m
1000 mm
cos 0°
cos 0°
Answer:
ℎ = 5.0 cm
Example 2-6 [Çengel, Y.A., Cimbala J.M. Fluid Mechanics: Fundamentals and Applications - 3rd Edition. 2014.]
Reference:
Yunus A. Çengel, John M. Cimbala.
Fluid Mechanics: Fundamentals and Applications - 3rd Edition.
1221 Avenue of the Americas, New York, NY 10020: McGraw-Hill, 2014.