4 Fundamental Definitions: System and Property Basic Unit...
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ES206 Fluid Mechanics UNIT A: Fundamental Concepts ROAD MAP . . . A - 1: Engineering Fundamentals A - 2: Fluid Properties and Characteristics Unit A - 1: List of Subjects What is Fluid Mechanics? Basic Dimensions in Engineering Basic Unit System Fundamental Definitions: System and Property
Transcript of 4 Fundamental Definitions: System and Property Basic Unit...
ES206 Fluid Mechanics
UNIT A: Fundamental Concepts
ROAD MAP . . .
A-1: Engineering Fundamentals
A-2: Fluid Properties and Characteristics
ES206 Fluid Mechanics
Unit A-1: List of Subjects
What is Fluid Mechanics?
Basic Dimensions in Engineering
Basic Unit System
Fundamental Definitions: System and Property
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What is Fluid? What is Fluid Mechanics?
• A fluid is defined as a substance that deforms continuously when acted by shear
forces
• Fluid mechanics is the engineering science that deals with the action of forces
and associated motion of fluids
Continuum Mechanics
• In classic fluid mechanics, “fluids” are treated as a hypothetical continuous
substance, called continuum
• Continuum assumption employs an idea that fluids are made up continuously
with many (many) small particles. A continuum is a body that can be sub-
divided into many (many) infinitesimal elements with properties being those of
the bulk material.
Advanced Subjects in Fluid Mechanics
• Hydrodynamics: flow of fluids (liquids) in no change of density
(incompressible) flows
• Gas Dynamics: flow of fluids (gases) that undergo significant change of density
(compressible) flows ➔ Aerodynamics: flow of air
Unit A-1Page 1 of 6
What is Fluid Mechanics?
Continuum assumptionLiquids – water, oil, . . .Gases – vapor, air, . . .
Hydrodynamics, Gas Dynamics, Aerodynamics
Basic Dimensions
• An engineering property can be described in terms of three basic dimensions
Mass (M) / Length (L) / Time (T) = MLT System
Force (F) / Length (L) / Time (T) = FLT System
• An engineering property (units) are all combinations of these three basic
dimensions (plus temperature: )
• So, is it possible to combine certain properties to create a “dimensionless”
property? YES, and it is very important in engineering!
For example . . . Reynolds number (Re):
( )( )( )
( )
( )( )( )
( )
3 1
1 1
Density Velocity LengthRe ?
Viscosity
ML LT L
ML T
− −
− −= = =
Unit A-1Page 2 of 6
Basic Dimensions in Engineering
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SI Unit System
Mass = Kilogram (kg) / Length = Meter (m) / Time = Second (s)
Temperature = Kelvin (K) / Celsius (C)
Force = Newton (N): 1 N = (1 kg)(1 m/s2)
Temperature unit conversion: K = C + 273
Common Problem (KILOGRAM “FORCE”)
Is 1 kilogram force = 1 kilogram mass ?
“An object of 1 kilogram MASS has WEIGHT (force) of 1 kilogram on
EARTH”
Kilogram “Force” (kgf) and kilogram “Mass” (kg) are different properties, and
cannot be mixed-up
Force unit conversion: 9.8 Newton = 1 kgf
Unit A-1Page 3 of 6
Basic Unit System (1)
SI (Systemé International) Unit System
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US Customary Unit System
Mass = Slug / Length = Feet (ft) / Time = Second (s)
Temperature = Rankine (R) / Fahrenheit (F)
Force = Pound (lb): 1 lb = (1 slug)(1 ft/s2)
Temperature unit conversion: R = F + 460
Common Problem (POUND “MASS”)
Is 1 pound force = 1 pound mass ?
“An object of 1 pound MASS has WEIGHT (force) of 1 pound on EARTH”
Pound “Force” (lb) and Pound “Mass” (lbm) are different properties, and cannot
be mixed-up
Mass unit conversion: 1 slug = 32.2 lbm
Unit A-1Page 4 of 6
Basic Unit System (2)
Traditional (U.S. Customary) = British Gravitational (BG) or English Engineering (EE) Unit System
Slug or Pound Mass (lbm)?
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Mass and weight are two different properties.
Mass = fundamental physical property (how much “matters” are packed into an
object: note that this is, in fact, “non-measurable” quantity . . .)
Weight = size of the gravitational pull (the “force” exerted on an object): this can
be measured, certainly.
Newton’s 2nd law states: Weight = Mass Gravitational Acceleration (W mg= )
Hence, 2
300 N30.6 kg
9.8 m/s
Wm
g= = =
Note: this mass is a unique property of an object (will never change); therefore,
under the gravitational acceleration of 4 ft/s, the weight of the same object becomes
( )( )2 0.3048 m30.6 kg 4 ft/s
1 ftW mg= = = 37.3 N
Unit A-1Page 5 of 6
Class Example Problem
Related Subjects . . . “Basic Unit System”
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System and Property
• System: an object of focus, enclosed by system boundaries
• Property: characteristics that can be measured
Extensive and Intensive Properties
• Extensive properties (B): depends on the extent of the system (area, weight,
length, mass, . . .)
Mass (m)
Linear Momentum (mV )
Energy (E)
• Intensive properties (b): independent of the extent of the system (density,
temperature, . . .)
B mb= or B
bm
=
B = Mass (m) ➔ 1B
bm
= =
B = Linear Momentum (m V ) ➔ m
bm
= =V
V
B = Energy (E) ➔ E
b em
= =
Unit A-1Page 6 of 6
Fundamental Definitions:System and Property
Extensive and Intensive Properties
ES206 Fluid Mechanics
UNIT A: Fundamental Concepts
ROAD MAP . . .
A-1: Engineering Fundamentals
A-2: Fluid Properties and Characteristics
ES206 Fluid Mechanics
Unit A-2: List of Subjects
Fluid Properties
Ideal Gas Model
Compressibility of Fluid
Vapor Pressure
Surface Tension
Viscosity
Newtonian and Non-Newtonian Fluids
Dynamic Viscosity of Common Fluids
Fluid Density
• Density (or “Mass” Density)
= mass per unit volume (kg/m3 or slug/ft3)
• Specific Weight (or “Weight” Density)
= weight per unit volume (N/m3 or lb/ft3)
• Specific weight and density are related: g =
Density of water: = 1,000 kg/m3 or 1.94 slug/ft3
Specific weight of water: = 9,800 N/m3 or 62.4 lb/ft3
Specific Gravity
• Specific Gravity (SG) is the ratio of the density (or specific weight) of a given
fluid to the water
fluid fluid
water water
or SG
=
• Specific gravity indicates “how heavy” the liquid is, compared against water:
SGHg = 13.6
Unit A-2Page 1 of 12
Fluid Properties
g =
Specific Gravity (SG)
Equation of State
• Ideal Gas Equation of State: p RT=
• R is the specific gas constant (air: 287 J/kgK or 1,716 ftlb/slugR)
Note) specific gas constant depends on the type of gas (gas specific):
Helium: 2,077 J/kgK or 12,420 ftlb/slugR
Hydrogen: 4,124 J/kgK or 24,660 ftlb/slugR
Nitrogen: 296.8 J/kgK or 1,775 ftlb/slugR
Oxygen: 259.8 J/kgK or 1,554 ftlb/slugR
• Pressure (p) must be the absolute pressure (not “gage” pressure)
Note) conversion of pressures:
pabsolute = pgage + patm
• Temperature (T) must be in absolute scale (not “relative scale”)
Note) conversion of temperatures:
K = C + 273 and R = F + 460
Unit A-2Page 2 of 12
Ideal Gas Model
RTp =
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Let us assume Nitrogen as an ideal gas. From Table 1.8 (textbook),
Nitrogen 296.8 J/(kg K)R =
Note) local standard atmosphere depends on altitude:
➔ Prescott (5,000 ft): 12.228 psi
➔ Daytona Beach (sea-level): 14.696 psi
Applying the equation of state (ideal gas law):
( ) ( )3 o1.5 kg/m 296.8 J/(kg K) 25 C 273 Kp RT = = +
= 132,670 Pa (132.67 kPa)
Note: this is the “absolute” pressure. The purpose of this problem is to determine
the “gage” pressure.
The gage pressure is the pressure above (or below, if vacuum) the atmospheric
pressure, where the given atmospheric pressure here is: atm 97 kPap =
Therefore:
gage abs atm 132.67 kPa 97 kPap p p= − = − = 35.67 kPa
Note that the atmospheric pressure is equal to “ZERO GAGE” pressure:
atm 0p = (gage)
Unit A-2Page 3 of 12
Class Example Problem
Related Subjects . . . “Ideal Gas Model”
Compressibility of Liquids
• Bulk Modulus of Elasticity (Ev) indicates “compressibility” of liquids:
Water: Ev = 2.2 GN/m2 (almost “incompressible”)
Compressibility of Gases
• Mach Number (M) indicates “compressibility” of gases: V
Mc
=
• Speed of Sound (c) of an ideal gas: c kRT= (k = 1.4 for air)
Expansion and Compression of Gases (Thermodynamic Processes)
• Isobalic process (the process is under the constant pressure): constantp =
• Isothermal process (the process is under the constant temperature):
constantp
=
• Isentropic process of gases (the process is frictionless and no heat exchange
with surroundings): “reversible” / “no heat exchange”
1.4constant
k
p p
= =
Unit A-2Page 4 of 12
Compressibility of Fluid
d
dp
VdV
dpEv =−= constant=p
constant=kp
pEv =
kpEv =
kRTc =
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The speed of sound for an ideal gas can be calculated by: c kRT= .
This equation is, however, cannot be applied for liquids (near incompressible).
Extremely high bulk modulus of elasticity of liquids leads to considerably high
speed of sound.
The speed of sound of a liquid can be determined by: vEc
= .
Bulk modulus of elasticity can be found from Table 1.6 (textbook).
For (a) gasoline: 9 2
3
1.3 10 N/m
680 kg/mc
= = 1,382.67 m/s = 3,092.94 mph
For (b) mercury: 10 2
4 3
2.85 10 N/m
1.36 10 kg/mc
=
= 1,447.61 m/s = 3,238.21 mph
For (c) seawater: 9 2
3 3
2.34 10 N/m
1.03 10 kg/mc
=
= 1,507.26 m/s = 3,371.64 mph
In case of air (at sea-level): 760 mphc =
Unit A-2Page 5 of 12
Class Example Problem
Related Subjects . . . “Compressibility of Fluid”
Vapor Pressure and Cavitation
• The pressure at which a change of phase (liquid => gas) occurs is called vapor
pressure
• Water boils at 212 F at 14.7 psi at sea-level
• Alternatively, vapor pressure is raised to 14.7 psi by increasing the temperature
to 212 F
• Water boiling occurs at a room temperature, if the pressure of the water is
reduced to its vapor pressure. This is called cavitation
• Therefore, the property “vapor pressure” of a liquid is not a constant value:
rather, it depends on the temperature.
For example: the vapor pressure of water is 1.77103 Pa (absolute) at 15.6 C.
However, you can also alternatively say that the vapor pressure of water is
exactly 1 atm (sea-level) at 100 C (this is called the “boiling”).
Unit A-2Page 6 of 12
Vapor Pressure
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Surface Tension
• Each portion of the liquid surface exerts tension on adjacent portions of the solid
surface that are in contact. This is called surface tension
• Surface tension () for a water-air interface is: = 0.0734 N/m
Capillary Motion
Liquid in a glass tube: for wetting liquid (for example, water), the liquid will rise
up in the glass tube (figure a). For non-wetting liquid (for example, Mercury), the
liquid will be depressed (figure c)
Applying the equation of equilibrium (figure b),
( 0yF+
= ➔ ( )2 2 cosR h R = )
2 cos
hR
= ( = 0 for water, = 130 for Mercury)
Unit A-2Page 7 of 12
Surface Tension
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Due to the surface tension, a capillary action (raising or depressing of a column of
liquid within a tube) will occur. The height of capillary action can be determined
by: 2 cos
hR
=
Properties of mercury can be found from Table 1.6 (textbook).
Mercury is a “non-wetting” liquid ( o130 = ): therefore,
( )( )( )
1 o
3 3 3
2 4.66 10 N/m cos130
133 10 N/m 1.5 10 mh
−
−
=
= −3.003×10−3 m (3.003 mm)
Negative sign means the column is depressed.
Unit A-2Page 8 of 12
Class Example Problem
Related Subjects . . . “Surface Tension”
Shear Stress in a Fluid
• Shear stress of a viscous fluid is proportional to rate of shear strain: du
dy =
du/dy represent “how much velocity change” is expected in the direction
perpendicular to the solid surface. The “velocity” can be interpreted as “how
much deformation of a fluid” in the direction parallel to the solid surface (shear
direction) is happening (shear strain).
• The proportional constant () is called dynamic (or absolute) viscosity
Viscosity
• Dynamic (or Absolute) viscosity (): units in Ns/m2 or lbs/ft2
• Kinematic viscosity ():
= (units in m2/s or ft2/s)
Unit A-2Page 9 of 12
Viscosity
dy
du = : Dynamic or
Absolute Viscosity
du
dy
Newtonian and Non-Newtonian Fluids
• Newtonian fluids: shear stress and rate of shear strain are directly proportional
du
dy = (Newtonian fluid)
• Non-Newtonian fluids: shear stress and rate of shear strain are not directly
proportional: Bingham plastic, shear thinning / thickening
Non-Newtonian fluids shows non-linear behavior (relationship between shear
stress and rate of shear strain cannot be described like Newtonian fluid). Non-
Newtonian fluids are not the main scope of this course: ES206 Fluid Mechanics
is a Newtonian Fluid Mechanics.
Examples of Non-Newtonian fluids include:
Shear thinning
Shear thickening
Bingham plastic
Unit A-2Page 10 of 12
Newtonian and Non-Newtonian Fluids
dy
du =
Viscosity of Gases and Liquids
• Viscosity is a function of temperature
• Empirical equation of viscosity for gases: Southerland’s equation: 3 2CT
T S =
+
For air: C = 1.45810−6 kg
m s K
and S = 110.4 (K)
• Empirical equation of viscosity for liquids: Andrade’s equation:
B TDe =
• Empirical equations (often also called, “correlations”) are the equations obtained
by typically experimental data based “curve-fit,” thus the equation is not based
on theory (based on the “fact”).
Unit A-2Page 11 of 12
Dynamic Viscosity of Common Fluids
ST
CT
+=
2/3
TBDe /=
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For a Newtonian fluid, the shear stress is related to the viscosity and the rate of
shear strain, such that: du
dy =
For fluid 1, the shear stress developed on the top plate is:
( ) 2 21top plate 1
top plate
3 m/s 2 m/s0.4 N s /m 20 N/m
0.02 m
du
dy
− = = =
For fluid 2, the shear stress developed on the bottom plate is:
( ) 2 22bottom plate 2
bottom plate
2 m/s 0 m/s0.2 N s /m 20 N/m
0.02 m
du
dy
− = = =
Therefore, the ratio of these two shear stresses are equal to 1
Unit A-2Page 12 of 12
Class Example Problem
Related Subjects . . . “Viscosity”
