What is Aerodynamics? A- 3: Governing Equations of...
Transcript of What is Aerodynamics? A- 3: Governing Equations of...
AE301 Aerodynamics I
UNIT A: Fundamental Concepts
ROAD MAP . . .
A-1: Engineering Fundamentals Review
A-2: Standard Atmosphere
A-3: Governing Equations of Aerodynamics
A-4: Airspeed Measurements
A-5: Aerodynamic Forces and Moments
AE301 Aerodynamics I
Unit A-1: List of Subjects
What is Aerodynamics?
Engineering Units Review
Flow Field Representations
Flow Field Properties
Equation of State
WHAT IS AERODYNAMICS?
Aerodynamics is a branch (means: a specific application) of Fluid Mechanics:
(i) Underlying assumption is the fluid as a continuum and,
(ii) The working fluid is standard air.
CLASSIFICATION OF AERODYNAMICS
1. Density: incompressible (constant density) or compressible
2. Viscosity: inviscid (zero viscousity) or viscous
=> If viscous: laminar, transitional, or turbulent
3. Velocity (Mach number): subsonic (M < 0.7), transonic (0.7 < M < 1), supersonic (1 < M < 5), or
hypersonic (5 < M)
=> If subsonic: incompressible (M < 0.3) or compressible (0.3 < M)
I. Analytical Solution:
=> Potential Flow Analysis (2-D & 3-D)
=> Panel Methods (2-D & 3-D)
=> Thin Airfoil (2-D) / Lifting Line & Surface Methods (3-D)
II. Experiment: Wind Tunnel (scale-model tests) / Prototype (full-scale tests)
III. Numerical Simulation: Computational Fluid Dynamics (CFD)
Unit A-1Page 1 of 8
What is Aerodynamics?
1. Density
2. Viscosity
3. Velocity
A. Space (1-D, 2-D, or 3-D)
I. Analytical Solution
II. Experiment
III. Numerical Simulation
B. Time (Steady, Periodic, or Unsteady)
CONSISTENT AND “INCONSISTENT” UNITS
“Consistent” (or coherent) units allow physical relationships to be written without the need of
conversion factors in the basic formulas.
“Inconsistent” units cannot be used against consistent units (important): inconsistent units must
be converted into consistent units, before executing numerical calculation.
SI UNIT SYSTEM
Kilogram or kg (mass), Meter or m (length), Second or s (time)
=> Force: Newton (N): 1 N = (1 kg)(1 m/s2)
“Inconsistent” force unit: Kilogram-Force: kgf (“Kilogram-Mass: kgm = kg” is consistent unit)
Unit conversion: 9.8 N = 1 kgf
(Do you casually use “Newtons” as a unit of force, or weight . . .?)
US CUSTOMARY (ENGLISH) UNIT SYSTEM:
Slug (mass), foot or ft (length), Second or s (time)
=> Force: Pound (lb): 1 lb = (1 slug)(1 ft/s2)
“Inconsistent” mass unit: Pound-Mass: lbm (note: “Pound-Force: lbf = lb” is consistent unit)
Unit conversion: 1 slug = 32.2 lbm
(Do you casually use “slugs” as a unit of mass . . .?)
Unit A-1Page 2 of 8
Engineering Units Review (1)
1 slug = 32.2 lbm9.8 N = 1 kgf
“Consistent” Unit:Slug / N
“Inconsistent” Unit:lbm / kgf
Both measures “weight” = 15 pounds force / 85 kilogram force
Mass is not a measurable property (you can never directly determine the amount of mass, by any
measurement method) . . . one can only “estimate” the mass, by measuring the force (weight)
generated by the presence of mass, under a known gravitational acceleration (W mg ).
WHAT IS THE BOTTOM LINE?
Under the gravitational acceleration on earth (9.8 m/s2 or 32.2 ft/s2):
A substance of 15 pounds mass “weighs” 15 pounds force.
A substance of 85 kilogram mass “weighs” 85 kilogram force.
Problems:
Pound mass (lbm) is “inconsistent” mass unit, and cannot be used against pound force (lb).
Kilogram force (kgf) is “inconsistent” unit, and cannot be used against kilogram mass (kg).
SO, “WHAT DO YOU NEED TO DO” IN ENGINEERING ANALYSIS?
In engineering, you should re-state “more accurately” as follows:
A substance of 0.466 slugs (15 lbm / 32.2 ft/s2 = 0.466 slugs) “weighs” 15 pounds force.
215 lb 0.466 slugs 32.2 ft/sW mg and 32.2 lbm
0.466 slugs 15 lbm1 slug
“Pound (force) can only be created by the combination of slugs, ft, and second)”
“Pound (force) and pound (mass) cannot coexist within the same equation”
“Under the standard sea-level gravity on earth (32.2 ft/s2), 1 lbm of object “weighs” 1 lbf”
A substance of 85 kilograms “weighs” 833 Newtons (85 kgf 9.8 m/s2 = 833 N).
2833 N 85 kg 9.8 m/sW mg and 1 kgf
833 N 85 kgf9.8 N
“Newton (force) can only be created by the combination of kilogram, meter, and second)”
“Kilogram (force) and kilogram (mass) cannot coexist within the same equation”
“Under the standard sea-level gravity on earth (9.8 m/s2), 1 kg of object “weighs” 1 kgf”
Unit A-1Page 3 of 8
Class Example Problem A-1-1
Related Subjects . . . “Engineering Units Review”
Suppose, if you go to “FRY’S” and measure a bunch
of fruits . . . The scale says, “15 pounds.” Does this
mean: 15 pounds mass, or 15 pounds force?
A standard health meter in Japan indicates everything
in “kilograms.” Suppose, if you step onto a Japanese
health meter and it indicates, “85 kilograms.” Does
this mean: 85 kilogram force, or 85 kilogram mass?
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STANDARD UNITS FOR AERODYNAMIC PROPERTIES
Pressure (SI Unit): N/m2 (Pascal, or “Pa”), 1,000 Pa = 1 kPa
Pressure (US Customary Unit): lb/in2 (psi), lb/ft2 (psf)
Pressure unit conversion: 1 psi = 144 psf
Density: kg/m3 (SI Unit) / slugs/ft3 (US Customary Unit)
Temperature (SI Unit): Celsius (C), Kelvin (K) / K = C + 273
Temperature (US Customary Unit): Fahrenheit (F), Rankine (R) / R = F + 460
Gas Constant (Standard Air): 287 J/(kgK), 1,716 (ftlb)/(slugR)
COMMON MISHAPS OF UNITS
Angle of Attack: may be given in degrees or radians.
Temperature: the problem may be given in relative scales (C / F), but usually need to use absolute
scales (K / R) in calculations.
Pressure: you need to know either “absolute” or “gage” pressure.
Do you use “inches” in your calculation?
Unit A-1Page 4 of 8
Engineering Units Review (2)
0 F = 460 R0 C = 273 K
1 atm (Standard Sea-Level) = 2,116 lb/ft2
1 atm (Standard Sea-Level) = 1.01105 N/m2
1 ft = 0.3048 m1 lb = 4.448 N
1 mile = 5,280 ft1 h = 3,600 s
60 mph = 88 ft/s
STREAKLINE AND PATHLINE
A streakline is the line defined in a flow field by continuously injecting infinitesimally small
particles at a fixed location. The idea of streakline is the pure experimental method (Eulerian).
A pathline is the line defined by tracing a path of an individual particle. A pathline can be visualized
by injecting small particles in the flow field. The idea of pathline is a particle trace (Lagrangian).
FLOW VELOCITY AND STREAMLINE
For a steady flow, a moving fluid element traces out a fixed path in space, called a streamline.
Streamlines are mathematical concept. Hence, streamlines can be formed by “equations of
streamlines” in a given flow field.
A streamline is a line, at which the velocity vector is tangent along its path.
NEWTONIAN FLUID SHEAR STRESS
Due to the viscosity, the shear stress will be developed between the adjacent streamlines.
For Newtonian fluids: dV
dy (: absolute or dynamic viscosity).
Unit A-1Page 5 of 8
Flow Field Representations
FLOW FIELD PROPERTIES AND STANDARD UNITS
Common properties in aerodynamics that can define flow field at a point:
(1) Pressure:
p = p (x,y,z,t): units => lb/ft2 (psf), lb/in2 (psi), N/m2, Pa
Pressure in absolue value: psia, psfa
Pressure in gage value: psig, psfg
(2) Density / Specific Volume:
= (x,y,z,t): units => slugs/ft3, kg/m3
Specific Volume: 1v (volume per unit mass: inverse of density)
(3) Temperature:
T = T (x,y,z,t): units => °F, °C, °R, K
Absolute Scale Temperature (conversion):
°R = °F + 460 / K = °C + 273
VELOCITY (A VECTOR PROPERTY)
(4) Velocity:
( , , , )x y z tV V : vector property (= magnitude + direction)
Vector Notation: publication / handwritten V VV
Unit Vectors: ˆ ˆ ˆi j k (Cartesian) / ˆ ˆ ˆ r ze e e (cylindrical)
Unit A-1Page 6 of 8
Flow Field Properties
in "absolute" pressurepsia psi
in "gage" pressurepsig psi
1 (Specific Volume)v
EQUATION OF STATE
A perfect (ideal) gas is the one in which the intermolecular forces are negligible.
Specific (R) and universal ( ) gas constants for a standard air:
Note 1): Pressure must be in absolute pressure (not “gage” pressure).
Note 2): Temperature must be in absolute scale (K / R, not C / F).
IDEAL GAS MODELS IN AERODYNAMICS
Rule of Thumb (means “usually,” but not always true)
Calorically Perfect => Thermally Perfect => Non Ideal Gas
Ideal Gas of Air Ideal Gas of Air of Air
Subsonic Supersonic Hypersonic
The difference between thermally and calorically perfect gases will be discussed later!)
Unit A-1Page 7 of 8
Equation of State
= 8,314 J/(kg mole K)
= 4.97 104 (ft lb)/(slug mole ºR)
M = 28.96 kg/(kg mole) or slug/(slug mole)
“Calorically Perfect” Ideal Gas of Air“Thermally Perfect” Ideal Gas of Air“Non-Ideal” Gas of Air
The pressure of air on the wing = 3.25 psi (144) = 468 lb/ft2
The density of air on the wing: 0.00075 slugs/ft3
Using the equation of state: p RT or p
TR
Note 1) Temperature must be in absolute scale temperature (R), not relative scale (F)
Note 2) Pressure must be in absolute pressure (not gage pressure)
2
3
468 lb/ft363.63 R
(0.00075 slugs/ft )[1,716 (ft lb)/(slug R)]
pT
R
In Fahrenheit: 363.63 460 = o96.37 F
Unit A-1Page 8 of 8
Class Example Problem A-1-2
Related Subjects . . . “Equation of State”
The air is flowing on a wing of Boeing 787 Dreamliner at
cruise. The pressure and density of air measured were 3.25
psi (absolute) and 0.00075 slugs/ft3, respectively. What is the
temperature (in F) of the air at that point?
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