PATTS REVIEW CENTER PATTS COLLEGE OF AERONAUTICS Lombos Ave., San Isidro, Parañaque
FUNDAMENTALS OF AERODYNAMICS Reviewer: Engr. R. Renigen A. THE INTERNATIONAL STANDARD ATMOSPHERE The atmosphere is the mechanical mixture of gases surrounding the earth. Atmospheric Constituents Nitrogen ------------------ 78.03% Oxygen ------------------- 20.99% Aragon --------------------- 0.94% Carbon Oxide ------------- 0.03% Hydrogen ------------------ 0.01% Helium --------------------- 0.004% Neon ----------------------- 0.0012% and a small amount of water vapor and other gases Four (4) Layers of Earth’s Atmosphere 1. 2. 3. 4.
Troposphere Stratosphere Ionosphere Exosphere
STANDARD VALUES FOR AIR AT SEA LEVEL Pressure PO
= 14.7 lb/in2 = 2116.8 lb/ft2 = 29.92” Hg = 76 cm Hg = 760 mm Hg = 101325 Pa = 1 atm
Density ρ0
= 0.002377 slug/ft3 = 1.225 kg/m3
Temperature: TO
= 519 °R = 288 K
Coefficient of Dynamic Viscosity μo
= 3.7372 x 10-7 slug/ft-sec = 1.7894 x 10-5 kg/m-sec
SEA LEVEL UP TO TROPOPAUSE
1
1. TEMPERATURE VARIATION WITH ALTITUDE T = TO + ah ;
θ=
T TO
Where: θ=
T , temperature ratio T0
T = temperature at any altitude above sea level up to tropopause in oR or K T0 = 519 °R or 288 K a = temperature gradient or lapse rate (= -0.009566 °R/ft or -0.00651 K/m or -6.51 K/km) h = any altitude above sea level up to tropopause in ft or m or km
2. PRESSURE VARIATION WITH ALTITUDE 5.26
P ah δ= = 1+ P0 T0
Where: δ=
P , pressure ratio P0
P = pressure at any altitude above sea level up to tropopause in lb / ft2 or Pa Po = 2116.8 lb/ft2 or 101325 Pa a = -0.003566 °R/ft or -0.00651 K/m or -6.51 K/km h = any altitude above sea level up to tropopause in ft or m or km
3.
DENSITY VARIATION WITH ALTITUDE 4.26
σ=
ρ ah = 1+ ρ0 T0
σ=
ρ , density ratio ρ0
Where:
ρ = density at any altitude above sea level up to tropopause in slug/ft3 or kg/m3 ρ0 = 0.002377 slug/ft3 or 1.225kg/m3 a = -0.003566 °R/ft or – 0.00651 K/m or -0.51 K/km h = any altitude above sea level up to tropopause in ft or m or km
ABOVE TROPOPAUSE (11 KM) up to STRATOPAUSE (32 KM)
2
1. TEMPERATURE VARIATION WITH ALTITUDE T = 390.15 °R or 216.5 K (constant from 11km up to 32 km) 2. PRESSURE VARIATION WITH ALTITUDE English System δ=
P 1.26 = 4.805 x10 −5 h P0 e
Where: P = pressure at any altitude above tropopause up to stratopause in lb / ft2 P0 = 2116.8 lb/ft2 h = any altitude above trotopause up to stratopause in feet Metric System δ=
P 1.26 = 1.578 x10 −4 h P0 e
Where: P = pressure at any altitude above tropopause up to stratopause in Pa Po = 101325 Pa h = any altitude above tropopause up to stratopause in meters 3. DENSITY VARIATION WITH ALTITUDE English System σ=
ρ 1.68 = 4.805 x10 −5 h ρ0 e
Where: ρ = density at any altitude above tropopause up to stratopause in slug /ft3 ρ0 = 0.002377 slug / ft3 h = any altitude above tropopause up to stratopause in feet Metric System σ=
ρ 1.68 = 1.578 x10 −4 h ρ0 e
Where: ρ = density at any altitude above tropopause up to stratopause in kg / m3 ρ0 = 1.225 kg/m3 h = any altitude above tropopause up to stratopause in meters ALTIMETERS
3
An altimeter is a pressure gauge which indicates an altitude in the standard atmosphere corresponding to the measured pressure. Pressure altitude, hp – is the altitude given by an altimeter set to 29.92 “Hg. Density altitude, hd – is the altitude corresponding to a given density in the standard atmosphere. Temperature altitude, hp – is the altitude corresponding to a given temperature in the standard atmosphere.
Problems: 1. Calculate the pressure, density and temperature at 8 km and 12 km altitudes in the standard atmosphere. 2. On a hot day, the measured temperature and pressure are 38°C and 29.0 “Hg, respectively. Calculate the density and the density ratio. 3. A standard altimeter reads 4500 meters when the ambient temperature is 275K. What is the density altitude and the temperature altitude? 4. At a certain altitude, a standard altimeter reads 3000 meters. If the density altitude is 25000 meters, find the true temperature. B. BASIC AERODYNAMIC PRINCIPLES AND APPLICATIONS 1. CONTINUITY EQUATION – Conservation of mass along a streamtube, such as in a wind tunnel.
where: °
m = mass flow of fluid in slug/sec or kg/sec
ρ = density of fluid in slug/ft3 or kg / m3 A = cross-sectional area of tube in ft2 or m2 V = velocity of fluid in ft/s or m/s
PRINCIPLE OF MASS CONSERVATION (LAW OF CONTINUITY)
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The mass flow of fluid that passing to one section of the tube in one second is equal to the mass flow of fluid that passing to the other section of the tube in one second. (a) For incompressible fluid, ρ = constant (M < 0.3 approximately) A1V1 = A2V2 AV = constant Differential form: dV V
=
−
dA A
(b) For compressible fluid, ρ ≠ constant (M ≥ 0.3 approximately) A1V1 = A2 V2 ρAV = constant Differential form: dV V
2.
+
dA A
+
dρ ρ
=
0
BERNOULLI EQUATION – Conservation of energy along a streamline. BERNOULLI’S PRINCIPLE In a continuous flow of liquid, as velocity increases, pressure decreases; and as velocity decreases, pressure increases. (a) For incompressible fluid, ρ = constant (M < 0.3 approximately) V2 2
V1 2
+
P ρ
+
P1 V2 = ρ 2
2
=
cons tan t 2
+
P2 ρ
The isentropic equation of state is not needed Application: Definition of equivalent airspeed Ve: Ve =
2(Pt − P ) ρ0
Where: Pt = total pressure or stagnation pressure P = static pressure ρo = density at sea level Definition of true airspeed V: Ve =
2(Pt − P ) ρ0
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or V
=
Ve σ
Where: ρ = density at altitude σ=
ρ , density ratio ρ0
(b) For compressible fluid, ρ ≠ constant (M ≥ 0.3 approximately) V2 γ P + = constant 2 γ −1 ρ
2
V2 γ P Vt γ Pt + = + 2 γ −1 ρ 2 γ − 1 ρt
or M2 =
γ−1 2 Pt γ + 1 −1 γ −1 P
The isentropic equation of state can be used. P ργ
=
P1 ρ1
P2 P1
γ
=
=
constant
P2 ρ2
γ
ρ2
γ
ρ1
γ
The speed of sound is given by: Va =
γP ρ
English System Va = γgRT
where: γ = 1.4
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R = 53.342 ft/ °R Va =49.02
T
where: Va = speed of sound in air in ft/s T = absolute temperature in °R Metric System Va =20.05
T
where: Va = speed of sound in air in m/s T = absolute temperature in K Application:
Definition of calibrated airspeed Vc: Vc
2
γ−1 Pt −P λ = +1 −1 γ −1 P0
2Va0
2
THERMODYNAMIC PARAMETER RELATIONSHIP γ
γ
γ
Va T2 γ−1 W2 P2 ρ2 = = = = 2 Va P1 ρ1 T1 W1 1
2γ
γ−1
Problems: 1. A circular pipe 50 meters long papers from 3m to 2m in diameter. What is the velocity per meter run at the bigger section. If the velocity there is 30 meters/sec, and if the fluid is passing from bigger section to the smaller section, is the velocity increasing or decreasing? 2. A water pipe 1m in diameter gradually tapers down to 0.5m, the rate of flow is 0.063m 3/sec. If the pressure is 12.5 KPa where the diameter is 1m, what is the pressure where the diameter is 0.5m? 3. An airplane is flying at sea level, the difference between total and static pressure is 1723 Pa. What is the airspeed in meters per second? 4. Air having the standard sea level density has a velocity of 30 m/s at a section of a wind tunnel. At another section having an area half as that at the first section the flow velocity is 178.8 m/s. What is the density at the second section? 5. The wind tunnel shown in Fig. Below, has a smallest section (test section) measuring 1.22 m by 1.0m, and a largest section of 4 square meters at a certain tunnel speed the manometer reading is 0.72m. The manometer liquid has a specific gravity of 0.85. Calculate the airspeed in the test section. Assume incompressible flow and standard sea-level conditions.
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6. In an undisturbed airstream the pressure is 101325 Pa, the density is 1.225 kg/m 3, and the velocity is 450 m/s. What is the pressure if the velocity is 190 m/s? 7. Air standard pressure and temperature, has density of 1.225 kg/m 3 . If the air is compressed adiabatically to 3 atm, what are the specific weight, density and the temperature. 8. What is the speed of sound in air at 3.500m altitude. 9. A horizontal pipe line enlarges from a diameter of 0.152 m at point A to a diameter of 0.305 m at point 3. The flow of water is 0.567 cubic meter per second and the pressure at point A is 68.93 KPa. What is the pressure at point B. 10. Alcohol (sp. Gr. = 0.80) is flowing through horizontal pipe which is 0.254 m in diameter with a velocity of 12.2 m/s. At a smaller section of the pipe, there is 41.5 KPa less pressure. Assuming that the flow is smooth. What is the diameter there? 11. An airplane is flying at sea level at an airspeed of 160 knots. What is the difference between total and static pressure? 12. The diameter at section (1) is 0.3m. The diameter at section (2) is 0.15m. What is the flow rate of a substance (sp. gr. = 0.85) if the pressure between sections (1) and (2) is 12.7 cm Hg? 13. Consider a venture with a throat-to-inlet area ratio of 0.8 mounted in a flow at standard sea level conditions. If the pressure between the inlet and the throat is 335 Pa, calculate the Velocity of the flow at the inlet. 14. Consider a low-speed subsonic wind tunnel with a 12/1 contraction ratio for the nozzle. If the flow in the test section is at a standard sea level conditions with a velocity of 50 m /s, calculate the height difference in a U-tube mercury manometer with one side connected to the puzzle inlet and the other to the test section. PHg = 1.36 x 103 kg/ m3. 15. A pitot static tube is used to measure the airspeed at the test section of a wind tunnel. If the pressure difference across the pilot-static tube is 0.11 m of water, what is the airspeed at the test section? If the ratio of the cross-sectional area between the largest section and the test section is 100:1, what is the airspeed at the largest section? Assume incompressible flow standard sea level condition. 16. Consider water flowing through a smooth pipe whose diameter is decreasing. At one location, the diameter is 12 cm. If the velocity there is 10 m/s. (a) Find the mass flow rate (b) At a station further down the pipe, the diameter is 4cm. Find the velocity at this section. 17. At the entrance of a natural gas pipeline, where the diameter is 1.32 m, gas at 251 KPa, absolute and 27oC at 18.2. At the exit of the pipe, where the diameter is 2.8m, the gas is at 132 KPa, absolute and 17oC. Solve symbolically for the velocity at the exit. Assume compressible flow.
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Venturi Tube The Venturi tube is a convergent –divergent tube with a short cylindrical throat or constricted section. This device determines the rate of flow of fluid through the tube by measuring the difference in pressure between the throat section and the entrance section.
By Incompressible Bernoulli Equation:
P1 ρ
2
V1 2
+
P2 ρ
=
V2 2
+
(
1 2 P1 −P2 = ρ V2 − V12 2
2
)
By Incompressible Continuity Equation: Q = A1V1 = A2V2 Q V = A1
2
2 1
Q V = A2
2
2 2
;
Then, P1 − P2 =
ρ Q 2 A 2
2 2 Q − A 1
P1 − P2 =
ρ 2 1 Q A 2 2
2 2 1 − A 1
Where: Q A P ρ
= = = =
rate of flow in ft3/s or m3/s cross-sectional area in ft2 or m2 pressure in psf or Pa density in slug/ft3 or kg/m3
9
Pressure: 1. A venture tube narrows down from 4 in. in diameter to 2 in. in diameter. What is the rate of water if the pressure at the throat i.e., 2 lb/in2 less than at the larger section? 2. A venture tube in 6 in. in diameter at the entrance, where the pressure is 10 lb/in 2 (gage). The throat is 4 in. in diameter, there the pressure is 6 lb/in 2 (gage). What is the flow of water? 3. A 12 in. by 6 in venture meter is located in a horizontal water line. If the pressure gages read 30lb/in2 and 16lb/in2, what is the flow rate? 4.
KUTA –JOUKOWSKI THEOREM – relation between force acting on a body and vortex circulation. Flow about a circular cylinder
V = 2V∞sinθ Where: V = tangential velocity in ft/s or m/s V∞ = freestream velocity in ft/s or m/s Ө = angle through the point on the surface of the cylinder with the main direction of the airflow in deg. By Incompressible Bernoulli equation: P ρ∞
+
V2 2
=
P∞ ρ∞
+
V∞ 2
2
P = P∞ +
1 ρ∞ ( V∞2 − V 2 ) 2
P = P∞ +
1 2 ρ∞ V∞2 − ( 2V∞ sin θ) 2
P = P∞ +
ρ∞ V∞ (1 − 4 sin 2θ) 2
[
]
2
Where: P = Pressure at any point on the surface of a circular cylinder in psf or Pa
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P∞ =
Freestream pressure in psf or Pa
ρ ∞=
Freestream density in slug/ft3 or kg/m3
Problems: 1. A uniform current of air with a speed of 100ft per sec. flows around a circular cylinder. At a distance from the cylinder the pressure is atmospheric. What is the pressure at a point on the surface of the cylinder so located that a radial line through the point makes an angle of 15o with the direction of airflow? 2. For the flow in the above problem, what is the pressure on the cylinder surface at a 90o arc from the direction of airflow? Lift due to Circulation Vortex - the circular circulation of particles of fluid having the same energy content and center of rotation.
V∞
V = 2V∞sinθ
V=
Γ 2π r
V = 2 V∞ sinθ +
dL = dF sinӨ dL = PrdӨ sinӨ dL = Pr sinӨ dӨ
V∞
11
Γ 2π r
By Incompressible Bernoulli Equation: P ρ∞
V2 2
+
P = P∞ +
P = P∞ +
P∞ ρ∞
=
+
ρ∞ V∞2 − V 2 2
(
V∞ 2
2
)
2 ρ∞ 2 Γ V∞ − 2V∞ sin θ+ 2 2Πr
2 Π ρ∞ 2 Γ dL = − P = P + V − 2 V sin θ + r sin θdθ ∞ ∞ ∞ ∫0 ∫ 2 2Πr −Π L
L =
P∞ρ∞ Γ
Where: L = Lift per unit span in lb/ft or N/m ρ∞ =
Freestream density in slug/ft3 or kg/m3
V∞ =
Freestream velocity in fps or mps
Γ = strength of circulation in ft2/s or m2/s ( = 2пrV) V = tangential velocity in fps or mps ( = 2пrN) r = radius of cylinder in ft or in N = rotational speed in rps
For total lift LT = L x l Where: l = length of the cylinder in ft or m Problems:
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1) A cylinder 30 in. in diameter rotates in an airstream of 70 mph. It develops 40lb. of lift per foot length. What is the rational speed? 2) A Cylinder 1.22 m in diameter and 3.5 at in length is rotating at 100rpm in an airstream of 18m/s. Determine the total lift and the strength of the circulation of the cylinder at SSLC. 5.
VISCOUSE EFFECTS , THE BOUNDARY LAYER AND FLOW SEPARATION SKIN FRICTION - skin friction is air resistance, and it is t he tangential component force on the surface of a boy due to the friction between the two particles. STREAMLINE AND TURBOLENT FLOW a streamline flow may be defined as smooth non-turbulent flow. A turbulent flow is defined as a flow characterized by turbulence, that is a flow in which the velocity varies erratically in both magnitude and direction with time. LAMINAR FLOW – the word laminar is derived from the Latin word lamina meaning a thin plate of metal or some other material. Laminar flow employs the concept that air is flowing in thin sheets or layers close to the surface of a wing with no disturbance between the layers of air. BOUNDARY LAYER – the boundary layer is that layer of air adjacent to the airfoil surface. The cause of the boundary layer is the friction between the surface of the wing and the air.
Illustration of Laminar and a Turbulent Boundary Layer LAMINAR BOUNDARY LAYER – Is the laminar boundary layers the flow is steady and smooth. As a result, the layer is very thin, and so the form drag is very small. Also, the velocity gradient at the watts, though large enough to give significant viscous stress, is yet only moderate, so that the skin friction, though not negligible, is also very small. δ=
5 .2 x R Nx
Where: δ = laminar boundary layer thickness x = transition point R N x = transition Reynolds number
The “rubbing” of the boundary layer on the flat plate give rise to friction forces of: friction drag. The skin-friction drag coefficient for one side of a plate in laminar flow is given by:
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Cf =
Df 1.328 = 2 1 RN 2 ρV S
Where: S = area of one side of plate RN = Reynolds number based on the total plate length TURBULENT BOUNDARY LAYER – in turbulent boundary layers the flow is unsteady and not smooth, but eddying. When the flow is transitioned to turbulent flow, the boundary layer thickness will be increased. In fact, this phenomenon is other used to determine the location of the transition region. The boundary layer thickness can be determined by:
δ=
0.37 x (R N ) 1 5
x
The skin-friction drag coefficient for a flat plate can be calculated with schichting’s formula:
Cf =
Df 0.455 = 2 1 ( log10 R N ) 2.58 2 ρV S
Where: ρ = air density in slug /ft3 or kg/m3 V = air velocity in ft/s or m/s L = characteristic length of the body in the flow direction and is equal to the chord length for an airfoil in feet or meters. µ = coefficient of dynamic viscosity in
slug ft − sec
or
kg in − sec
For air, µ increases with temperature and can be calculated by the following approximate formula for the standard atmosphere: 3
2.329 x10 −8 T 2 µ= T + 216
,
slug , T in °R ft − sec
or 3
1.458 x10 −6 T 2 µ= T + 110.4
,
kg , T in K m − sec
At low Reynolds numbers, the flow in the boundary layer is laminar (laminar boundary layer). Above certain “transition Reynolds numbers”, the flow becomes turbulent (turbulent boundary layer) Transition take place on a flat plate at point x determined by:
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(R )
N x crit .
ρVx = µ crit .
= 3.5 x105
to 10 6
Problems: 1) Two plats, one having 6 ft span and 3 ft chord, the other having 9ft span and 6 ft. chord, are placed in different airstreams. The freestream velocity for the smaller plate is 100ft/sec. It is found that the total skin-friction drag for the two plates is the same. Find the airspeed for the larger plate. Assume laminar flow at standard sea level conditions. 2) Consider the stabilizer on a light airplane as a flat for the purpose of determining its skin-friction drag. If the transition RN is 750,000, what is the skin-friction drag of a rectangular stabilizer having a span of 1. 83 m and a chord of 0.91 in at a speed of 44.7 meters per second? Assume standard sea level conditions. 3) An airplane is flying at a density altitude of 4500 in at an ambient temperature of 234K. If the wing chord is 1.83 in and the equivalent airspeed is 103 meter per second, what is the overall Reynolds number of the wing? WIND TUNNEL A device for testing aircraft and its force components in a controlled airstream under laboratory conditions. TYPES OF LOW – SPEED WIND TUNNELS 1. Open –circuit tunnel
2. Closed circuit tunnel
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Forces and motion of airplane under testing a. lift b. drag c. side force
d. pitching moment e. yawing moment f. rolling moment
Importance of RN Used in comparison of flow pattern of on theorem bodies which are geometrically similar but not in dimension. Flow pattern similarity at a particular point 1. magnitude of velocity at constant proportion 2. direction of flow is the same 3. both bodies must be oriented or positioned in similar altitudes 4. both bodies must be positioned at the same eagle of attack Problems: 1) Find RN, for an airplane wing, 4 ft chord, moving at 130 mph through standard atmosphere. 2) Find RN for an airplane wing, 4 ft-chord, moving at 150 mph, air is 40oC; barometer reading is 29 m. Hg. 3) Find the velocity at which test should be run in a wind tunnel on a model wing of 0.10 m chord in order that RN, shall be the same as for a wing with a 1.22 m chord at 44.7 m/s. Air under standard conditions in both cases. 4) In a variable – density wind tunnel, under what pressure should test be run on a model with a 3-in chord, air velocity being 60 mph, in order that the RN shall be the same as for a full-size wing, of 4ft chord, moving at 100 mph through the air? Air temperature is the same in each case.
Variable Density Wind Tunnel
- a wind tunnel in which the air density can be increased by means of compressed air.
Flat Plates a] Flat plates perpendicular to airstream F
= 1.28
ρ AV 2 2
Where:
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F = Force on a flat plate normal to airstream in lb or N ρ = freestream density in slug/ft3 or kg/m2 A = cross-sectional area of the ft2 or m2 V = freestream velocity in fps or mps Problems: 1. What is the total force of a 45-mph wind on a hangar door 40 ft by 25ft? 2. An automobile windshield is 40 in. wide by 15 in. high and is vertical. What is the force against the windshield at 60 mph. 3. What is the force against the side of a building 70ft long and 40 ft high in a 90 – mph wind? 4. What force is required to push a flat plate 3 ft. by 2 ft, at a speed of 35 ft/s in a direction perpendicular to its surface?
Curved Deflecting Surfaces
= pAV2 ( 1 – cosЄ)
F = ρ AV 2 sin ∈ 2
=
F
FH + FV
2
F = ρ AV 2( 1− cos∈ ) 2
Where: FH = horizontal component of force F in lb or N FV = vertical component of force F in lb or N F = resultant force in lb or N V = freestream velocity
∈
= angle of deflection in deg.
A = cross –sectional area of airstream in ft3 or m2
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Problems: 1. A stream of air 60 ft. wide and 8 ft. high is moving horizontally at a speed of 100 mph. What force is required to deflect it downward 10o without loss in speed. 2. A stream of air 60 ft wide and 8 ft high is moving horizontally at a speed of 75mph. What force is required to deflect it downward 10o without loss in speed. 3. A stream of air 9.30 square meter in cross section is moving horizontally at a speed of 67 mps. It strikes tangentially against the interior wall of a semi circular cylinder so that it is deflected through 180o. What is the total force against the cylinder.
AIRFOIL THEORY An airfoil is a streamlined body which when set at a suitable angle of attack, produces more lift than drag.
DEFINITION OF AIRFOILGEOMETRY
DEFINITION OF AIRFOIL GEOMETRY MEAN CAMBER LINE – is the line joining the midpoints between the upper and lower surfaces of an airfoil and measured normal to the mean camber line. CHORD LINE – is the line joining the end points of the mean camber line THICKNESS - is the height of profile measured normal to the chord line. THICKNESS RATIO – is the maximum thickness to the chord ratio,
t . c
CAMBER – is the maximum distance of the mean camber line from the chord line. LEADING – EDGE RADIUS – is the radius of a circle, tangent to the upper and lower surfaces, with its center located on a tangent to the mean camber line drawn through the leading edge of this line.
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AERODYNAMIC FORCES AND MOMENTS ON AN AIRFOIL
DEFINITION OF SECTION FORCES AND MOMENT
FACTORS AFFECTING THE AERODYNAMIC FORCE (f): 1. 2. 3. 4. 5.
Velocity of air, V Air density, p Characteristic area or size, S Coefficient of dynamic viscosity, µ Speed of sound (compressibility effect), Va
FORMULAS : LIFT FORCE l = Cl
1 ρV 2 c 2
DRAG FORCE d = Cd
1 ρV 2 c = C d q c 2
PITCHING MOVEMENT m = Cm
1 ρV 2 c 2 = C m q c 2 2
where: 1 = lift force in N/m d = drag force in N/m m = pitching moment in N.m/m
IMPORTANT AIRFOIL CHARACTERISTICS The following relationships are of fundamental importance to airplane design and airplane analysis:
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LIFT CURVE
:
Cl versus α
The linear portion of the lift curve can be represented mathematically by the equation: Cl
=
a (α – α0)
or C lα (α – α0)
Cl = DRAG POLAR
:
Cl versus Cd
An important parameter representing the aerodynamic efficiency of airfoil is Cl the lift-drag ratio. The maximum magnitude can be obtained by drawing a tangent to the cd – cl Cd PITCHING MOMENT CURVE
:
Cm versus α or Cm versus Cl
The magnitude of cm depends on the location of the moment center. One important moment center which is often used is the so-called aerodynamic center (a.c.) It is defined as the point about which the moment coefficient is independent of α.
GEOMETRY FOR FINDING THE AERODYNAMIC CENTER
AIRFOIL PRESSURE DISTRIBUTION The pressure distribution is normally expressed in terms of the pressure coefficient, Cp.
Cp
=
P − P∞ q
=
P − P∞ 1 2 ρ ∞ V∞ 2
At low speeds, according to the incompressible Bernoulli Equation, 2
Cp
V = 1 − V ∞
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CRITICAL VELOCITY, Vcr 1
Vcr
=
Va ∞
( γ −1) M ∞2 + 2 2 γ +1
FOR AIR, γ = 1.4 1
Vcr
=
Va ∞
M ∞2 + 2 2 6
CRITICAL PRESSURE, Pcr 1
Pcr
( γ −1) M ∞ 2 + 2 2 = P∞ γ +1
FOR AIR, γ = 1.4 1
M 2 + 5 2 = P∞ ∞ 6
Pcr
CRITICAL PRESSURE COEFFICIENT, C p C pcr
=
Pcr − P∞ q∞
=
2 2 γM ∞
cr
γ
C p cr
2 + ( γ −1) M ∞2 γ−1 −1 γ +1
FOR AIR, γ = 1.4 C p cr
=
5 + M ∞2 6 2 0.7 M ∞
3.5
− 1
PROBLEMS: 1. An airplane is flying at an altitude of 3500m at an airspeed of 300 m/s. Find the critical speed, critical pressure and critical pressure coefficient. 2. What is the critical value of the pressure coefficient for an airplane flying at 500 knots in air at 25oF. 3. For an airplane flying at 270 mps at 25, 0000ft. attitude, find the critical value of the pressure coefficient.
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4. An airfoil has a lift-curve slope of 6.3 per radian and angle of zero lift of -2o. At what angle of attack will the airfoil develop a lift of 140lb/ft at 100mph under standard sea level conditions? Assume c = 8ft.
DESIGN OF AIRFOIL To design an airfoil for any specific use, the following effects of airfoil geometry should be noted: 1. Camber shape will affect mainly αp and cm . Any increase in camber will make ap and cm more negative. 2. Thickness distribution will change the value of lift curve slope, a.c. location, and center of pressure location. 3. The leading-edge shape has a pronounced effect on stall characteristics and on NACA AIRFOILS DESIGNATION 4 – digits airfoils :
Example NACA 4412
4 = Camber 0.04c 4 = Position of the camber at 0.4c from L.E. 12 = Maximum thickness 0.12c 5 –digit airfoils : Example NACA 23012 2 = camber 0.02c = design lift coefficient is 0.15 times the first digit for this series. 30 = Position of camber at
0.30 c = 0.15c 2
12 = Maximum thickness 0.12c 6 –digit airfoils
: Example NACA 65 -421
6 = Series designation 5 = Minimum pressure at 0.5c 3 = The drag coefficient is near its minimum value over a range of lift coefficient of 0.3 above and below the design lift coefficient. 4 = design lift coefficient 0.4 21 = maximum thickness 0.21c 7- series airfoil : Example NACA 747A315 7 = series designation 4 = favorable pressure gradient on the upper surface from L.E. to 0.7c at the design lift coefficient. A = a serial letter to distinguish different sections having the same numerical designation but different mean line or thickness distribution. 3 = design lift coefficient 0.3 15 = maximum thickness 0.15c
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AIRFOIL SELECTION In selecting an airfoil for an airplane lifting surface (wing, tail, or canard) the following considerations are important: 1. 2. 3. 4. 5. 6.
Drag (for example: to obtain the highest possible cruise speed) Lift –to-drag ratio a values of cl important to airplane performance Thickness (to obtain the lowest possible structural weight) Thickness distribution (to obtain favorable span loading and/or high fuel volume) Stall characteristics (to obtain gentle stall characteristics) Drag-rise behavior (associated with item 1)
GEOMETRIC FACTORS AFFECTING AIRFOIL MAXIMUM LIFT AT LOW SPEEDS The main features of airfoil design which affect wing stall and hence, the maximum lift coefficient, are: 1. THICKNESS RATIO For a given thickness ratio,
t , C l max very much depends on the leading-edge radius. It seen that c
the new NASA LS = (Low speed airfoils, a thickness ratio of about 13% will produce the best volume of maximum lift. For the newer LS airfoils the maximum lift occurs at around 15% thickness. 2. LEADING EDGE RADIUS Clmax depends not only on the thickness ratio, but also on the ratio of section thickness at 5% chord to the maximum thickness
t . c
The ratio
t is indicative of leading–edge radius. c
Therefore, a relatively large leading-edge radius is beneficial to producing large C l speeds.
max
at low
3. CAMBER AND LOCATION OF MAXIMUM THICKNESS The addition of camber is always beneficial to C l max and the benefit grow with increasing camber. The increment to maximum lift due to camber is least for sections with relatively large leading edge radius (i.e., the benefit of camber grows with reducing
t ; and camber is more c
effective on thin sections than on thick sections). In addition, a forward position of maximum camber produces higher values of C l For example, the NACA 23012 airfoil (with 2% maximum camber at 0.15 chord) has a C l max of 1.79 as compared with 1.67 for NACA 4412 (with 4% camber at 0.4 chord but the same thickness distribution) at a Reynolds number of 9 x 106 max
4. REYNOLDS NUMBER For airfoils with moderate thickness ratio, there is a significant because increase in C l with increasing Reynolds number. On the other hand, for thin airfoils the effect of Reynolds number is relatively significant. In general, these Reynolds number effects are less for cambered than for symmetrical sections. At low Reynolds number, the effect of camber is more insignificant. The opposite is true at Reynolds number greater than 6 x 106, where camber losses some of its effects max
EFFECT OF HIGH LIFT DEVICES ON AIRFOIL MAXIMUM LIFT TAILING –EDGE FLAPS
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Plain Flaps 1. Formed by hinging the rear-most par of the wing section about within the contour. 2. The main effect produced by the flap deflection is an increase in the effective camber of the wing. 3. The optimum flap angle is approximately 0.25. 4. The optimum flap angle is approximately 60o. 5. Leakage through gap 1330th of the chord resulted on a loss of 0.35 in C l
max
.
6. The maximum achievable increment is C l max is approximately 0.9. Split Flaps 1. The unusual split flaps is formed by deflecting the aft portion of the lower surface about a hinge point on the surfaces of the forward edge of the deflected portion. 2.
The optimum flap chord ratio is approximately 0.3 for 12% thick airfoils, increasing to 0.4 or higher for thicker airfoils.
3. The optimum flap angle is approximately 70o 4. The maximum achievable increment in C l
max
is approximately 0.9.
5. The optimum thickness ratio is approximately 18%. Slotted Flaps 1. Slotted flaps provide one or more slots between the main portion of the wing section and the deflected flap. 2. The optimum flap chord ratio is approximately 0.9 3. The optimum flap angle is approximately 40o for single slots and 70o for double-slotted flaps. 4. The optimum thickness ratio is approximately 16% 5. The maximum achievable increment in C l max is approximately 1.5 for single slots and 1.9 for double slotted flaps. Fowler Flaps The Fowler Flaps uses the same principles as the slotted flap, except that the flap also moves backwards in addition to a downward deflection. Thus, the effective wing area is increased! LEADING –EDGE DEVICES Slats Leading –edge slats are airfoils mounted ahead of the leading edge of the wing such as to assist in turning the air around the leading edge at high angle of attack and thus delay leading-edge stalling. They may be either fixed in position of retractable. The use of slats may increase C l by as much as 0.5. max
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Slots When the slots is located near the leading edge, the configuration differs only in detail from the leading edge slat. Additional slots may be introduced at various chordwise stations. The effectiveness of the slot derives from its BLC (= Boundary Layer control) effect. At low angle of attack, the minimum profile drag may be greatly increased with such slots. Leading –Edge Flaps A leading flap may be formed by bending down the forward portion of the wing section to form a droop. Other types of leading –edge flaps are formed by extending a surface downward and forward from the vicinity of the leading edge (Kruger flap) Leading Edge flaps reduce the severity of the pressure peak ordinarily associated with high angle of attack and thereby delay separation.
BOUNDARY LAYER CONTROL Higher maximum lift coefficient can also be achieved by boundary layer control (BLC). The idea may involve injecting high-speed for parallel to the wall (called “blowing”), removing the low energy boundary layer flow by “suction”, or both. Blowing is done to re-energize the boundary layer flow to dlay the separation, while suction is equivalent to eliminating the low energy shear layer.
25 VARIOUS CONFIGURATIONS OF TRAILING-EDGE FLAPS VARIOUS CONFIGURATIONS OF LEADING-EDGE DEVICES
