Basic Aircraft.engineers.pvsbico.aerodynamicsd

8/17/2019 Basic Aircraft.engineers.pvsbico.aerodynamicsd

http://slidepdf.com/reader/full/basic-aircraftengineerspvsbicoaerodynamicsd 1/350

A1, Page 1

FLIGHT

OPERATIONS

ENGINEERING

Aerodynamics (1): The Basics



A1, Page 2

Aerodynamics (1): The Basics

• Physical quantities

• Units

• Conversions

• Newton’s Laws



A1, Page 3

Mass

• A measure of the amount of matter in an object.

• Is a measure of an object’s inertia; exhibited by itsresistance to acceleration (change in velocity).

• Is independent of location - mass is constant.

• Units:

Metric system: kilograms

English system: slugs

• Conversion:

Kilogram mass = slugs mass ÷ 0.068522



A1, Page 4

Time

• Units:

– All systems: seconds, minutes, hours, days,weeks, fortnights

• Conversions:

minutes = seconds ÷ 60

hours = minutes ÷ 60

= seconds ÷ 3600



A1, Page 5

Temperature

• ‘Temperature’, the degree of hotness of an object, isa measure of the average molecular motion in a gas.

• Temperature is a measure of the averagetranslational kinetic energy associated with therandom, microscopic motion of molecules.

• Heat flows from higher temperature regions to lowertemperature regions.

• Two scales (actually, systems) of temperature are incommon use:

‘Relative’ temperature

‘Absolute’ temperature


http://slidepdf.com/reader/full/basic-aircraftengineerspvsbicoaerodynamicsd 6/350 A1, Page 6

Temperature

Relative temperature is a scale on which two points areestablished - usually the freezing and boiling points of

water - and to which arbitrary values are assigned.Once these points are established, the scale is set:

32 degrees and 212 degrees on Fahrenheit scale

(Difference = 180 degrees)

0 degrees and 100 degrees on Celsius scale

(Difference = 100 degrees)

change of 1 degree C = change of 1.8 degrees Fahrenheit



Temperature

• Absolute temperature is a scale of temperature in which

zero degrees represents the minimum temperaturetheoretically possible. In theory, at this temperature allmolecular motion ceases. Therefore, it is impossible fortemperature to go below this zero value.

• Absolute temperature is the scale that must be used inairplane performance equations.

• There are two absolute temperature scales in general

use: Rankine, and Kelvin.



Temperature

• Rankine scale:

Zero degrees is the minimum (occurs at -459.67 deg. F).Increment of temperature is the same as Fahrenheit scale,with 180 degrees between freezing and boiling points ofwater (freezing at 491.67, boiling at 671.67).

• Kelvin scale:

Zero degrees is the minimum (occurs at -273.15 deg. C).Increment of temperature is the same as Celsius scale, with

100 degrees between freezing and boiling points of water(freezing at 273.15, boiling at 373.15).



Temperature

• Units:

Metric system: degrees Celsius (Relative)

degrees Kelvin (Absolute)

English system: degrees Fahrenheit (Relative)

degrees Rankine (Absolute)

• Conversions:

°C = (°F - 32) * 5/9 °F = (°C * 9/5) + 32°K = °C + 273.15 °R = °F + 459.67

°R = (°K * 9) ÷ 5



°C °K °F °R

100

0

-273.15 0

273.15

373.15

-459.67

32

212

0

491.67

671.67water boilingpoint

freezing

point

Absolutezero

Metric English

Temperature



Question #1:

The local weather report in Seattle is forecasting a high

temperature for today of 59 degrees Fahrenheit

(accompanied, of course, by light rain). What is this

temperature in units of both Celsius and Kelvin?

Question #2:

I will be traveling to Dubai, United Arab Emirates, and am

told to expect high temperatures near 45 degrees Celsiuseach day of my trip. I’m more familiar with the English

system of temperature measurement, so 45 degrees does

not sound very warm to me. What is this temperature in

units of both Fahrenheit and Rankine?



Length or Distance

• Measurement of one-dimensional space

• Units:

Metric system: millimeters, centimeters,meters, kilometers

English system: inches, feet, yards, statute miles

Aeronautical/Nautical: nautical miles

length



Length or Distance

Conversions:

centimeter = millimeters ÷ 10

meters = centimeters ÷ 100kilometers = meters ÷ 1000

feet = inches ÷ 12.0

yards = feet ÷ 3.0

inches = centimeters ÷ 2.540

statute miles = feet ÷ 5280.0meters = feet ÷ 3.28084

nautical miles = meters ÷ 1852.0

nautical miles = feet ÷ 6076.1 = statute miles ÷ 1.1508



Question #3:

The runway length available at airport XYZ ispublished as 3025 meters. All of my takeoffperformance information has been provided

in units of feet. What is the runway lengthavailable at airport XYZ in units of feet?



Area

• Measurement of two-dimensional space

• Units:

Metric system: square meters, square centimeters

English system: square yards, square feet, square inches• Conversions:

square meters = square feet ÷ 10.7639

square inches = square centimeters ÷ 6.4516



Volume

• Measurement of three-dimensional space

• Units:

– Metric system: liters, cubic meters, cubic centimeters

– English system: U. S. gallons, imperial gallons, cubicfeet, cubic yards, cubic inches

– Aeronautical: liters, U. S. gallons, Imperial gallons



Volume

Conversions:

cubic meters = cubic feet ÷ 35.3147

= cubic yards ÷ 1.3079

U. S. Gallons = liters ÷ 3.7854

= cubic inches ÷ 231.0

Imperial gallons = U. S. Gallons ÷ 1.2009

= cubic inches ÷ 277.42



Question #4:

A 747-400 has been loaded with 50000gallons of fuel. How much fuel is this inunits of liters?



A1, Page 19

Velocity (Speed)

• Distance traveled per unit of time

• Units:

Metric system: kilometers per hour, meters per second

English system: miles per hour, feet per second

Aeronautical: knots (nautical miles per hour),feet per second, Mach number

d

velocity = distancetime



A1, Page 20

Velocity (Speed)

Conversions:

knots = miles per hour ÷ 1.1508

= kilometers per hour ÷ 1.8520

= feet per second ÷ 1.6878



A1, Page 21

Question #5:

For a given set of conditions (including sealevel, standard day, zero wind) a 777-200ER

has a recommended rotation speed ontakeoff of 156 knots. What is this speed inunits of feet per second?



A1, Page 22

Acceleration

• Change of velocity per unit of time =

• Units:

Metric system: meters per second per second, g’s

English system: feet per second per second, g’s

Aeronautical: knots per second, g’s

• Conversions:

meter/sec2 = ft/sec2 ÷ 3.28084

∆V

∆t



A1, Page 23

Question #6:

If an airplane accelerates down the runway at auniform rate from a velocity of 0 to a velocity of

260 feet per second in 25 seconds, what is itsaverage acceleration?



A1, Page 24

Mach Number

Ratio of an airplane’s true airspeed to the speed of sound for

the conditions at which the airplane is flying

(a discussion of the speed of sound appears in a later section)

True Airspeed

speed of soundMach = =

VTAS

a



A1, Page 25

Force

• The effort required to accelerate an element of mass

• Force equals mass times acceleration (F=ma)

• Units:

Metric system: Newtons, kilograms force

English system: pounds



A1, Page 26

Force

• Definitions:

– Newton is the force required to accelerate one kilogram ofmass at the rate of 1 meter per second per second.

– Kilogram (force) is the force required to accelerate onekilogram of mass at the rate of one standard ‘g’.

– Kilogram of force thus equals 9.80665 Newtons.

– Pound is the force required to accelerate one slug of massat the rate of one foot per second per second.

– Pound is also the force required to accelerate a mass of.45359237 kilograms at one standard ‘g’. It is thus equalto 0.45359237 kilograms force.



A1, Page 27

Force

Conversions:

kilograms (force) = pounds ÷ 2.2046226

= Newtons ÷ 9.80665

pounds = kilograms (force) ÷ 0.45359237

= Newtons ÷ 4.4482



A1, Page 28

Question #7:

An empty 757-300 is weighed, and theweight is recorded as 135000 pounds.What does this airplane weigh in unitsof kilograms?



A1, Page 29

Density

• Mass per unit volume (mass-density)

• Weight per unit volume (weight-density)

• Units:

Metric system: kilograms per cubic meter(kg/m3), grams per cubic centimeter

English system: slugs per cubic foot

Aeronautical: kilograms per liter, kilos per cubicmeter, pounds per gallon, poundsper cubic foot



A1, Page 30

Density

Conversions:

kg (force) per liter = pounds per gallon ÷ 8.3454

Water at 4 degrees Celsius has a density of exactly1 kg/liter ( = 1 gram per cubic centimeter, or 1000kg per cubic meter, or 62.430 pounds percubic foot)



A1, Page 31

Specific Gravity (Relative Density)

• Ratio of the density of a material compared to

density of water

• Aviation - used to specify fuel density (e.g. thespecific gravity of fuel is typically around 0.805)

• Specific gravity is approximately equal to thedensity of a material in kilograms per liter



A1, Page 32

Energy

Various forms of Energy:

– Work: force applied over a distance

– Kinetic: due to motion of an object (work an object can doby virtue of its motion)

– Potential: ‘stored’ mechanical energy (work an object can

do by virtue of its configuration)

– Internal: energy associated with the random, disorderedmotion of molecules on a microscopic scale

– Heat: energy transfer due to a temperature difference

– Chemical: stored energy released via a chemicalreaction (e.g., combustion)



A1, Page 33

Energy

Units:

Metric system: Joules (Newton-meters)calories (gram-degree C at 14.5 deg C)

English system: BTU (pound-deg F at 63 degrees F)Foot-pounds



A1, Page 34

Energy

Conversions:

Joules = calories ÷ 0.238846

Newton-meter = foot-pound ÷ 0.73757

BTU = foot-pound ÷ 778.15

= Joules ÷ 1055.1

= calories ÷ 252.02



A1, Page 35

Power

• Rate of work = = F * V

• Units:

Metric system: Watt (Newton-meter per second)

English system: foot-pounds per second, horsepower

• Conversions:

horsepower = foot-pounds per second ÷ 550

= Watts ÷ 746

F * ∆d

∆t



A1, Page 36

Question #8:

An unnamed person weighs 180 pounds.

It takes that person exactly one minute to climb the 72

steps from the parking lot to their office on the thirdfloor each morning. Each step is 6.75 inches high.

How many horsepower does this person produce

when climbing these stairs each morning? (neglectany forward motion)



A1, Page 37

Pressure

• Force per unit area

• Units:

Metric system: Newtons per square meter(Pascals), bars

English system: pounds per square foot

Aeronautical: atmospheres, hectoPascals

(millibars), inches of Mercury,millimeters of Mercury



A1, Page 38

Pressure

Conversions:

bars = Newtons per square meter ÷ 100,000= Pascals ÷ 100,000

hectoPascals (mb) = Pascals ÷ 100

Pascals = lb per square ft ÷ 0.208855atmospheres = mb ÷ 1013.25

= in Hg ÷ 29.92

= mm Hg ÷ 760.0

= lb per square ft ÷ 2116.2166



A1, Page 39

Atmosphericpressure

(Liquid Hg)

Height = 29.92 in.When pressure = 1 std atmos

Atmosphericpressure

Barometer:Used to indicate atmospheric

pressure based on a

measurement of the height

of Hg in an inverted tube

p=0

(p2 = po + ρgh)

Pressure



A1, Page 40

Question #9: An airplane is parked at a sea level airport where theoutside air pressure is currently 768 mm Hg. Standardday air pressure at this airport is 1013.25 millibars

(mb). In units of mb, how much higher or lower thanstandard day pressure is the current pressure?

Question #10: An airplane is parked at a sea level airport where theoutside air pressure is currently 29.68 inches Hg.Standard day air pressure at this airport is 1013.25

millibars (mb). In units of mb, how much higher orlower than standard day pressure is the currentpressure?



A1, Page 41

Angles

• Units:

All systems: degrees, minutes, seconds, radians

• Conversions:

radians = degrees * = degrees * .0174533

r

r

r

1 r a d i a n

• Radius laid out along the outlineof a circle covers an angle of

one radian.

• Circumference of circle = 2πr

• 360 degrees = 2π radians

180π



A1, Page 42

Useful Trigonometry

θ

xy

z

Tan θ =

Sin θ =

Cos θ =

zy

x

y

xz

x * Cos θ = z

x * Sin θ = y

Sin θ

Cos θ= Tan θ

θ = flt path angle

V

VHORIZ

VVERTV * Cos θ = VHORIZ

V * Sin θ = VVERT

For Example:



A1, Page 43

Newton’s Laws of Motion

1. Every object persists in its state of rest or uniform motion ina straight line unless it is compelled to change that state byforces impressed on it.

2. Force is equal to the change in momentum (mV) per changein time. (For a constant mass this is more commonly writtenas F= ma).

3. For every action, there is an equal and opposite re-action.



A1, Page 44

End of Aerodynamics (1):

The Basics



A2, Page 1

FLIGHT

OPERATIONS

ENGINEERING

Aerodynamics (2): Weight



A2, Page 2

Aerodynamics (2): Weight

• Weight• Gravitational Acceleration



A2, Page 3

Weight

• “Weight” is force - the force exerted on an object bygravitational attraction. It cannot be measured directlyhowever, instead, we infer weight by measuring theamount of force which is required to prevent an objectfrom accelerating due to gravitational attraction. Thatrestraining force will be equal and opposite to the forceof gravitational attraction.

• Weight is the product of gravitational accelerationand mass.

• Units of weight are force - Newtons, pounds,kilogram-force



A2, Page 4

Weight

• Mass is constant, but weight is not constant. This is dueto variations in gravitational acceleration.

• Gravitational acceleration varies with:

– Latitude – Altitude

– Ground Speed

– Direction of Motion• Reference/Standard value for gravitational acceleration:

go = 32.17405 ft/sec2

= 9.80665 m/sec2

Weight = mass * gravitational acceleration



A2, Page 5

Gravitational Acceleration

• “Gravitational attraction” (“gravity”) will cause two

masses to accelerate toward each other if notrestrained. That acceleration is called gravitationalacceleration. “Falling” is simply a term used todenote acceleration of an object toward the earth

due to gravitational attraction.

• In the absence of other forces this acceleration canbe predicted by the Universal Law of Gravitation



A2, Page 6

Acceleration of mass M1 toward M2: a = GM

2d2

center

of mass

center

of mass

Universal Law of Gravitation

Mass M2

Mass M1

Force of attraction: F = GM1M2

d2

d

F



A2, Page 7


• Masses exert equal and opposite forces of attraction on each other.• Attraction force depends upon product of the masses and the inverse

square of the distance between their centers of mass:

• Acceleration of M1 toward M2 depends upon mass of M2 (not M1) andthe square of the distance between their centers:

e.g., in the absence of factors such as aerodynamic drag, all objects

fall towards the earth with the same acceleration - a feather will fallas rapidly as a ball of lead.

Acceleration of mass M1 toward M2: a = G M2

d2

(G = gravitational constant = 3.44 * 10-8 ft2lb/slug2)

Force of attraction: F = G M1M2

d2



A2, Page 8


• Earth’s gravitational acceleration, per the Universal Law

of Gravitation, is a function of: – The earth’s mass

– Inverse square of the distance between the center of

the earth and the center of mass of an object.• Radius of the earth is not constant; depends on latitude.

North Pole



A2, Page 9

Equator

North Pole

South Pole

a = 20,925,780 ft

Radius of earth: r E =

a4 + b4 tan2(Ø)

a2 + b2 tan2(Ø)

r E

b =

2 0 ,

8 5 5 , 6

3 6

f t

Ø = latitude

(at sea level)



A2, Page 10

Question #1:

What is the sea level radius of the earth at

latitude 47.55 degrees north? (Latitude of PaineField in Everett Washington).



A2, Page 11

Variation with Latitude

gØ,SL = go [ 1 - 2.6373 x 10-3 Cos (2Ø) + 5.9 x 10-6 Cos2 (2Ø) ]

• Where: go = Reference gravitational acceleration

= 32.17405 ft/sec2 = 9.80665 m/ sec2

= gØ,SL at a latitude of 45 degreesØ = Latitude

• Accounts for: – Average density distribution of the earth

– ‘Oblateness’ of the earth

– Earth’s rotation

(Lambert’s Equation - Stationary Object Over a Rotating Earth)

Rotation rate of the



A2, Page 12

Equator

Rotation rate of theearth = ωE

Ø

r E

r E cosØ

f t o f

at i o

n



A2, Page 13

Equator

r E cosØ

r E


Centrifugal acceleration(due to earth’s rotation)

= r E cos Ø * ωE2

A c c e l d

u e

t o

g r a v i t y

Ø

Ø

C o m

p o n e

n t

c e n t r i f u g

a l a c c e l e r a

o p p o

s i n g g r a v i t y

r E c o

s 2 Ø

*

E 2



A2, Page 14

Variation with Latitude and Altitude

gØ,Z = [ gØ,SL + ωE2 r E Cos2(Ø) ] [ r E / (r E + z) ]2 - ωE

2(r E + z) Cos2(Ø)

Where: r E =

a = 20,925,780 ft

b = 20,855,636 ft

ωE = Rotation rate of the earth7.29212 x 10-5 radians per sec

z = Altitude (feet)

Ø = Latitude

a4 + b4 tan2(Ø)

a2 + b2 tan2(Ø)

(Stationary Object Over a Rotating Earth)

R i f h f ion



A2, Page 15

Equator

(r E + z) cosØ

r E

A c c e l d

u e

t o g r a v i t y

Ø

z

Ø


C o m

p o n e

n t o f

c e n t r i f u g a l

a c c e l e

r a t i o

o p p o

s i n g g

r a v i t y

Centrifugal acceleration(due to earth’s rotation)



A2, Page 16

Question #2a:

A 777-200ER is sitting on the runway at PaineField airport in Everett Washington, latitude =47.55 deg. N, elevation = 571 ft, and weighs650,000 lb. What is its mass in slugs?

Note: gØ,SL = g47.55 deg, SL = 32.18159 ft/s2

Cos2(47.55) = .45555



A2, Page 17

Question #2b:

If the same airplane were sitting on the runway atChangi airport in Singapore, latitude = 1.21 deg. N,elevation = 23 ft, what would be its weight?

Note: gØ,Z = g1.21 deg, 23 FT = 32.08939 ft/s2)



A2, Page 18

Question #2c:

How much less does this airplane weigh at Changi

airport compared to what it weighs at Paine Fieldairport (46.34 deg difference in latitude)?

V i ti ith G d S d



A2, Page 19

Variation with Ground Speed

Where: VG = Ground Speed (knots)

• Provides the change in gravitational acceleration asa function of ground speed.

• Airplane is flying in a curved path around the earth -the faster it flies, the greater the centrifugalacceleration.

∆ gCENTRIFUGAL =- ( 1. 6878099 * VG)2

(r E + z)

(Moving Object Over a Stationary Earth)

North Pole



A2, Page 20

Equator

North Pole

South Pole

r E

z

Equator

South Pole

VG

A c c e l d

u e

t o g r a v i t y

C o m p o

n e n t

o f

c e n t r i f u g

a l a c

c e l

o p p o

s i n g g r a v

i t y

Variation with Direction of Motion –‘C i li ’ C ti



A2, Page 21

‘Coriolis’ Correction

Where: TTR = True track angle (degrees)

• Adjusts the centrifugal acceleration for the interactionbetween the East-West component of the airplane’s groundspeed and the rotation of the earth.

• Correction is:

– Zero when flying north, south, or over the poles.

– Maximum when flying east or west at the equator.

∆ gCORIOLIS = -2 ωE * 1. 6878099 * VG * Cos Ø * Sin (TTR)

(Moving Object Over a Rotating Earth)

North Pole



A2, Page 22

South PoleCentrifugal accelerations for

ground speed and rotation ofthe earth have been applied

independently of each other,

and do not produce the correct

result of their combined effects.

The Coriolis correction

adjusts the centrifugalaccelerations to produce the

correct result of their

combined effects.

VG(E-W) + VωE

VG(E-W) + VωE

VG

VG(E-W)

VG(N-S)

VωE

Westbound

VG

VG(E-W)

VG(N-S)

VωE

Eastbound



A2, Page 23

Question #3a:

If the 777-200ER of question 2a is flying westbound (truetrack = 270 deg.) at a ground speed of 480 knots, alatitude of 47.55 deg. N, and an altitude of 37000 ft.,what is the lift required to offset the airplane weight?

Note: g = (gØ,Z + ∆gCENTRIFUGAL + ∆gCORIOLIS)

gØ,Z = g47.55 deg, 37000 FT = 32.067613 ft/s2



A2, Page 24

Question #3b:

If this same airplane, at the same mass, were flyingeastbound (true track = 90 deg.) at a ground speed of 480knots, a latitude of 1.21 deg., and an altitude of 37000 ft.,what is the lift required to offset the airplane weight?

Note: at 1.21 deg. N; 37,000 ft; Vground = 480 knots;

True Track = 90 deg.:

g1.21 deg, 37000 FT = 31.975693 ft/s2

∆gCENTRIFUGAL = -0.0313099 ft/s2

∆gCORIOLIS = -0.1181277 ft/s2



A2, Page 25

Question #3c

How much less lift is required for this same airplane

to fly under the conditions of 3b relative to 3a?

Summary



A2, Page 26

Summary

Weight = mass ∗‘net’ gravitational acceleration

= mass ∗ (gØ,Z + ∆gCENTRIFUGAL + ∆gCORIOLIS)

• ‘net’ gravitational acceleration is the result of: – Gravity (Universal Law of Gravitation)

– Centrifugal acceleration due to earth’s rotation

– Centrifugal acceleration due to airplane motion



A2, Page 27


Weight



A3, Page 1

FLIGHT

OPERATIONS

ENGINEERING

Aerodynamics (3): Air That Is Not Moving




A3, Page 2


• Air as a Fluid

• Viscosity

• Pressure

• Hydrostatic equation

• Equation of State

• Specific Heats

Air as a Fluid



A3, Page 3

Air as a Fluid

• Aerodynamic forces on an airplane are produced by the

airplane’s interaction with the air. Therefore, tounderstand airplane performance we need to knowsomething about the properties of air.

• Air is a fluid because it behaves in same manner as otherfluids. That is, air flows in a manner similar to fluids suchas water, and follows the same rules of behavior.

• "Aerodynamics" (predicting the way something will

behave in air) is the science of fluid dynamics dealingspecifically with air.

Ideal and Real Fluids



A3, Page 4

Ideal and Real Fluids

• Ideal fluids

– Incompressible – No viscosity

• Real

– Compressible

– Viscous

Viscosity



A3, Page 5

Viscosity

• Viscosity is the property of a fluid which creates

resistance to motion of an object through that fluid.• May be thought of as ‘stickiness’; the tendency of a fluid

to stick to the surface of an object it’s moving over.

• Viscosity makes a fluid tend to bend around an objectover which it is flowing.

• Air exhibits viscosity, although the magnitude is small.

• Viscosity is one of causes of aerodynamic drag.

Viscosity



A3, Page 6

Viscosity

For example:

If a metal ball is released in acontainer of very thick oil, it will at

first accelerate downward, but soonit will reach a “terminal velocity” as itfalls through the fluid. The greaterthe viscosity, the slower the terminalvelocity.

The terminal velocity is the speed atwhich the viscous resistance force just equals the weight of the sphere,resulting in zero net force and thuszero acceleration.

Vball

Pressure



A3, Page 7

Pressure

• The molecules of a gas are constantly in random motion.Due to this motion, the molecules of gas in a container

are constantly bouncing off the walls of the container. All of these countless, tiny, molecular collisions whenadded together exert an average force on the walls ofthe container.

• The rate of molecular motion depends on the temperatureof the gas. The higher the temperature, the greater therate of molecular motion, hence the greater the force.

• Pressure is the force exerted by these molecular impactsper unit of area - for example, pounds (force) per squareinch, etc.

Hydrostatic Equation



A3, Page 8


• Atmosphere is composed of gasses.

• Gasses have weight, however small.

• Picture a column of air, extending from the earth’ssurface to the upper limit of the atmosphere.

• Similar to any other fluid, for static balance the pressureof the air at any altitude must equal the pressure due tothe weight of the air in the column above it.




A3, Page 9


p r e s

s u r e =

p

p r e s s u

r e = p + d p

dh

Column of fluid (air)with cross sectional

area A

For a fluid at rest (i.e., not in motion):

dp = - ρg dh

The hydrostatic equation is derived merely by

equating the vertical forces on the block of fluid:

p * A = (p + dp) * A + ( * A * dh)g

upwar d force due to

pressur edownwar d force due to pr essure

downwar d force due to weight of

the block of fluid

Equation of State



A3, Page 10

Equation of State

This equation expresses the relationship betweenpressure, temperature and volume of a gas.

P = ρRT

Where: P = pressure

ρ = density (mass/volume)

R = a constant called the ‘specific gas constant’which varies with the type of gas

T = temperature (absolute units)

Equation of State



A3, Page 11

Equation of State

• At constant temperature (isothermal): Pressuretimes volume is constant. If volume doubles,pressure decreases by half.

• At constant volume (isochoric): Pressure andtemperature are directly dependent on each other.If temperature doubles, pressure will double.

• At constant pressure (isobaric): Volume andtemperature are directly dependent on each other.If temperature doubles, volume must double.

P = ρRT PV = mRT

Specific Heats of Air



A3, Page 12


• The specific heat of a material is the amount of heat

necessary to raise the temperature of a unit massof the material by one degree.

• For a gas, there are two distinct ways in which theheating operation may be performed which are ofparticular interest: at constant volume, or atconstant pressure.




A3, Page 13


The amount of heat required to raise a unit mass ofair one degree at constant volume (isochoric process)

is called CV .

ρ = constant

T P

Heat is added

area

For air: Cv = 4290 ft*lb/slug/oR (English

Units)




A3, Page 14


The amount of heat required to raise a unit mass ofair one degree at constant pressure (isobaric) is

called CP.

P = constant

T ρ

( F r i c t i o n l e

s s l i d )

Increase in volume

Heat is added

area

For air: CP = 6006 ft*lb/slug/oR

F

(English

Units)




A3, Page 15


• An important parameter in high-speed flows is the ratio ofspecific heats, denoted as gamma (γ ).

γ = Cp / Cv = 1.4 for air

• The gas constant, R, is derived from Cp and Cv as follows:

R = Cp - Cv = 1716 ft*lb/slug/oR for air

• Therefore, R is the amount of mechanical work which isobtained by heating a unit mass of the gas through a unittemperature rise.

(English

Units)



A3, Page 16


Air That Is Not Moving

FLIGHT



A4, Page 1

OPERATIONS

ENGINEERING

Aerodynamics (4):

Standard Atmosphere / Altitude Measurement

Aerodynamics (4):Standard Atmosphere/Altitude Measurement



A4, Page 2

p

• The Atmosphere

• Standard Atmosphere

• Altitude Measurement

• Non-standard pressureand temperature effects

The Atmosphere



A4, Page 3

p

• The performance of the airplane and engine depends on

the generation of forces by the interaction between theairplane or engine and the air mass through which it flies.It’s thus necessary to examine the properties of the earth’satmosphere; the thin layer of gas surrounding the earth.

• Earth’s atmosphere is a mixture of nitrogen (~78%), oxygen(~21%), argon (~0.9%), and small amount of other gases.Water vapor is always present but in varying amountsdepending on temperature and relative humidity (usuallyless than one percent at earth's surface).

The Atmosphere



A4, Page 4

p

• Energy of the sun is responsible for heating the earth’satmosphere, but most of this heating is done indirectly.Most of the energy goes into heating earth's surface,

which in turn heats the air.• Warm air near earth’s surface rises, expands, decreases

in pressure, and cools as altitude increases.

• Equilibrium is reached where no more reduction intemperature occurs. Altitude at which this occurs isnamed the “tropopause.”

– Tropopause: Altitude at which temperature reduction

ceases; this altitude varies with latitude. – Troposphere: Region from earth’s surface tothe tropopause.

– Stratosphere: Region above the tropopause.

Temperature is essentially constant in this region.

The Atmosphere



A4, Page 5

p

• The atmosphere is dynamic, constantly changing.Pressure, temperature and density of the air are affected

by: seasonal changes, moving air masses, latitude andlongitude, time of day, solar sunspot activity, etc.

• Therefore, we need a standard basis to provide commonreference conditions for estimating and comparing engine

and airplane performance, and for comparing flight test andwind tunnel results.

• Many different standard atmosphere models exist, but thestandard used in most of the world by airplane and enginemanufacturers is that established by the International Civil Aviation Organization (ICAO) back in the early 1960’s.

Standard Atmosphere



A4, Page 6

p

• International Standard Atmosphere (ISA)

– Model of the atmosphere determined by averagingdata gathered over a long period.

– Data collected mostly in the mid latitudes of thenorthern hemisphere, therefore the standard is most

representative of conditions in these regions. – Even though the expected deviations from this

standard may be much larger in polar or equatorialregions, the same standard is used as a reference.

• Before covering the details of the standard atmosphere,let’s cover a few altitude concepts leading up to it.

Absolute Altitude



A4, Page 7

• Physical distance from the center of the earth• “g” varies with this distance

• Required for space flight calculations, but not very

useful for airplane performance (except for precisecalculations of acceleration of gravity)

Ab s o l u t e Al t i t u d e

Tapeline Altitude (Geometric)



A4, Page 8

p ( )

• Physical distance above the sea-level surface of the earth.Distance that would be measured by a tape measure dropped

to sea level.

• Would like to know this height for terrain clearance.

• Cannot actually be measured with a pressure measuringdevice such as an altimeter, but can be modeled such that apressure measuring device can give a good indication of it.

(A device such as a radio altimeter would more correctly read

physical distance above the ground. But, would not want to flyto such an altimeter because displayed altitude would beconstantly changing depending on the terrain being flown over)

Tapeline Altitude (Geometric)



A4, Page 9

p ( )

Equator 20,925,780 ft

2 0 ,

8 5 5 ,

6 3 6

f t

htape

htape

Note: The same tapelinealtitude is a different absolute

altitude at each latitude

Model of Tapeline Altitude



A4, Page 10

Assumptions:

1. Air behaves as a perfect gas. Therefore, we can use the‘equation of state.’

2. Atmosphere is static with respect to the earth. Therefore, we canuse the ‘hydrostatic equation.’

3. Dividing equation 2 by equation 1, and integrating, we obtain:

p = ρRT

dp = - ρg dh

dp = - 1 g dh

p R T∫ ∫

1

2

Model of Tapeline Altitude



A4, Page 11

Generally, atmospheric models for pressure, temperature,and density as functions of altitude are created by:

– Defining T and p at sea level. – Assuming a particular variation of temperature with

altitude (based on experimental results).

– Calculating the corresponding variation of pressure withaltitude (using the equation above).

– Calculating the corresponding density using the ‘equationof state’.

dp = - 1 g dh

p R T∫ ∫

Geopotential Altitude Model



A4, Page 12

To simplify the calculations, ‘g’ is assumed to be

constant with altitude, equal to its standard sea levelvalue go. Using this assumption, it can be pulled outof the integral. The result of this is a model we call ageopotential model; a model that assumes a constant

value of g with altitude.

(Equation #1)dp = - go 1 dhgeo

p R T∫ ∫

Standard Atmosphere



A4, Page 13

There is a special geopotential model where

the variation of temperature with altitude isbased on the ICAO work mentioned previously.This special model we call the ‘InternationalStandard Atmosphere’ – the model used in all

airplane performance work.

Standard Atmosphere



A4, Page 14

The ‘International Standard Atmosphere’ is a specificgeopotential model of the atmosphere based on thefollowing “defined” temperature variation with altitude:

Altitude ~ ft

Temperature ~ oC

Sea level:15 oC

(59 oF)

36,089 ft and above:

-56.5 oC

(-69.7 oF) Lapse rate with altitude

from sea level to 36,089 ft:

- 1.9812 oC 1000/ft(- 3.566 oF 1000/ft)

15-56.5

36,089

0

Standard Atmosphere



A4, Page 15

• Standard temperature

Sea level: To = 15 °C (= 288.15 °K)

Below 36089 ft: TSTD = To - .0019812 * ∆ h (°C) Above 36089 ft: TSTD = -56.5 °C

• Temperatures are often provided in terms of an increment from

standard day temperature; e.g. std + 10 °C

• In airplane performance work we generally use the temperatureratio (theta):

θ =To

T T and To must be in

units of absolute temp



A4, Page 16

Standard Atmosphere



A4, Page 17

• Values of θ are tabulated for standard day only. They must beadjusted to obtain values for non-standard day.

• ‘ISA +’ and ‘Standard +’ mean the same thing. They mean thetemperature obtained from the standard atmosphere plus theincluded temperature increment.

• Example:

Altitude = 31,000 ft.

Temperature = ISA + 10 °C

What is the value of θ?

TSTD = 226.75 °K

To = 288.15 °Kθ = =

To

T 226.75 + 10

288.15= 0.8216

(Not .7869)

Standard Atmosphere



A4, Page 18

• Actual atmospheric pressure is rarely used in performancework. Instead, just as we defined a temperature ratio, θ, wealso define a pressure ratio, δ (delta).

• Using this definition of δ, along with the defined temperaturevariation with altitude, and equation #1, we obtain thevariation of δ with altitude.

δ =po

p Where po, standard pressure at sea level,is defined in the standard atmosphere as =

29.92126 in. Hg. (= 1013.25 millibars)

δ = (1-6.875586 * 10-6* hp)

5.25588 δ = .223360 * e36089 – hp

20805.7( )

Below 36,089 ft: Above 36,089 ft:

Standard Atmosphere



A4, Page 19

• The relationship between δ (pressure) and altitude inthe standard day model defines pressure altitude. Most

performance work uses pressure altitude, and this is thebasis for all altimeter calibrations.

• Values of δ tabulated in standard atmosphere tablescan be used even for non-std day.

• Notice that we used a temperature model to relate T toaltitude, and relations between p and T to relate p toaltitude. However, the way the data is used is

somewhat reversed; we know (or measure) p or δ, andcalculate altitude, called pressure altitude, hp. Thestandard atmosphere tables then provide the standardT for that pressure altitude.



A4, Page 20

Standard Atmosphere



A4, Page 21

• Just as with temperature and pressure, we define a density ratio,σ (sigma).

• Equations can be written for density, σ, as a function of altitude,

but it is easier to use the following relationship, which can bederived from the equation of state:

• As with θ, values of σ are tabulated for standard day only. Theymust be adjusted to obtain values for non-standard day. This canbe accomplished using the equation above.

δ = σ θ

σ = ρ o

ρ Where ρo, standard density at sea level,

is = .00237692 slug/ft3 = po

RTo

σ = δ

θ



A4, Page 22

Question #1a:

For a pressure altitude of 33,000 ft. and astandard day (i.e., ISA + 0 °C), calculate: δ, θ, σ



A4, Page 23

Question #1b:

For a pressure altitude of 33,000 ft. and atemperature of ISA + 10 °C, calculate: δ, θ, σ

Geopotential Altitude Model



A4, Page 24

• Assuming a constant value for ‘g’ in the model creates afictitious height called geopotential height, hgeo.

• Why is this called geopotential?The potential energy of an object raised through a change ingeopotential height of ∆hgeo in a constant gravitational field, g= g

o

, is the same as the potential energy of the same objectraised through a change in tapeline height of ∆htape in thevariable gravitational field.

mgo ∆hgeo = m g dhtape∫htape initial

htape final

(Additional Information)

Tapeline Altitude Versus Geopotential Altitude



A4, Page 25

• Relationship between altitude in a Geopotential model and

tapeline model (that is, between a constant ‘g’ model and avariable ‘g’ model):

Where: r = 20,855,531 ft. (average radius of the earth)

• htape is greater than hgeo at all altitudes greater than sea level

htape =hgeo * r

r - hgeo


Pressure Altitude Versus Geopotential Altitude

(Addi i l I f i )



A4, Page 26

How are pressure altitude and geopotential altitude related?

Altitude ~ ft

Temperature ~o

C

0

-∆T +∆T

S t a n d a r d d a y

S t d . +

S t d . -

In performance work we onlylook at temperature distributions

that differ by a constant

temperature increment from the

standard variation with altitude.



(Additi l I f ti )



A4, Page 27

Remember that the basis for all geopotential models (thestandard atmosphere falls into this category) is as follows:

For pressure altitude (std. Atmosphere):

For any other geopotential model withtemp distribution TSTD + ∆T:

dp = - go 1 dhgeo

p R T∫ ∫dp = - go p

dhp R TSTD

dp = - go p

dhgeo R (TSTD + ∆T)



(Additi l I f ti )



A4, Page 28

For small changes in altitude, the following can be used:

However, to find the relationship between pressure altitude andgeopotential altitude at any point in the atmosphere the followingshould be used:

∆hgeo = ∆hpTSTD + ∆T

TSTD

hgeo = hp -go

R ∆Thp lnδ


Altitude Measurement



A4, Page 29

• When calculating the performance of an airplane or engine,we need to know the altitude for which the performance is tobe computed, since airplane and engine performancedepends on air pressure, among other factors.

• It is difficult to measure the true height (“tapeline” or“geometric”) of an airplane above the ground , and even if itwere easy, true height is of no relevance to airplane

performance. Performance depends on air pressure, andknowing tapeline height does not tell us the air pressure, sincethe atmosphere is not the same from day to day.

• We could publish performance in terms of air pressuremeasured in lb/in2, or N/m2, or any other units. However,pilots think in terms of airplane height. It’s therefore moreuseful to express performance in terms of altitude.Considering atmospheric variations from day to day, how isthis to be accomplished?




A4, Page 30

• Consider a pressure gauge such as the “aneroid” pressuregauge; a gauge which doesn’t use a liquid. It works on theprinciple of differential pressure between the inside andoutside of a sealed chamber, one surface of which is

connected by a mechanism to a pointer.

• An increase in pressure on the outside relative to the insideof the chamber causes the pointer to deflect in one

direction; a decrease in pressure causes the pointer todeflect in the other direction. The position of the pointer maythen be calibrated in terms of the pressure existing outsidethe chamber (i.e., ambient pressure).

• An “aneroid barometer,” such as that shown on the nextpage, is simply an aneroid pressure gauge measuring theatmospheric pressure. It may be used to predict weatherconditions as they depend to a large degree on atmospheric

pressure and changes in that pressure.

ure



A4, Page 31

P r e s

s u r

D e c r e a

s e

P r e s s u r e I n c r e a s e

Flexible

Chamber

Walls

Flexible

ChamberWalls

Aneroid Barometer

Sealed

Air

Chamber




A4, Page 32

• We know from our atmosphere discussion that air pressure isdirectly related to height. Suppose, then, that we propose to installa pressure gauge in the airplane and to use it to indicate theairplane’s height. To use a pressure gauge to show height, we’ll

have to calibrate the gauge’s dial in units of height instead of unitsof pressure. We would then have an altimeter made from ananeroid pressure gauge.

• In order to calibrate a pressure gauge in units of height, it’snecessary to assume a relationship between pressure and height.

• The relationship we use to calibrate a pressure gauge in terms ofheight is the International Standard Atmosphere definition.

• “Pressure altitude” (or “pressure height”) is the heightcorresponding to the air pressure according to the ISA definition.




A4, Page 33

• All altimeters are calibrated to the pressure-height relationshipof the standard atmosphere. Therefore, altimeters sense thechange in pressure altitude.

• Remember from the standard atmosphere equations:

• These same equations are solved for altitude, as a function ofpressure ratio, δ:

Below 36,089 ft:

δ = (1-6.875586 * 10-6* hp)

5.25588 δ = .223360 * e36089 – hp

20805.7( )

Above 36,089 ft:

hp = 145,447 1- hp = 36089 -20805.7 ln

Below 36,089 ft (δ ≥ .223360): Above 36,089 ft (δ < .223360):

p

po

.190263

4.47708 p

po

A l t i t u

d e

re a s

e



A4, Page 34

The barometer is

recalibrated in terms ofaltitude based on the

standard day

relationship of pressure

versus altitude.

A I n

c r e

Pressure Altitude

0

Pressure 29.92

Al t i t u d e D e c r e a s e

Altitude



A4, Page 35

Question #2:

Using the Standard Atmosphere tables, findthe pressure altitude corresponding to an airpressure of 13.75 inches of Mercury.

Altimeter Errors Due to Non-Std Pressure



A4, Page 36

• One source of inaccuracy inherent in using a pressuregauge (calibrated in units of height) as an altimeter is the

fact that for any given elevation, atmospheric pressurechanges from day to day as weather patterns change. Thiswould cause an airplane’s altimeter, as the airplane sits onthe ground at some airport, to indicate different heights from

day to day.

• For takeoff and landing, it’s useful to the crew to know theirairplane’s height above the airport, particularly relating to the

local terrain. But, as mentioned above, the height indicatedby a calibrated pressure gauge will depend on theatmospheric conditions. Hence, we need a way to correct,or adjust, an altimeter for the effects of atmospheric

variations.




A4, Page 37

Altimeter indication usingstandard day calibration.

Elevation

29.92

28.86

27.82

26.82

25.84

4000 feet

3000 feet

2000 feet

1000 feet

Air Pressure (Inches of Mercury)




A4, Page 38


At a given elevation, thelocal air pressure will vary.

Elevation

29.92

H i g h - P

r e s s u

r e D a y

S t a n d a r d D

a

y

L o w - P r e s s u r

e D a y




A4, Page 39

Changing Air Pressure atConstant Elevation

At constant elevation, indicatedaltitude changes as local pressurechanges. This is unacceptable! Indicated

Altitude

29.92

S t an d ar d D a y

Hi gh -P r e s s ur eD a y

L ow-P r e s s ur eD a y




A4, Page 40

• To solve this problem we need some way to adjust thealtimeter calibration to compensate for variations in

atmospheric pressure.

• This adjustment is accomplished by using an“altimeter setting.” The altimeter setting allows the crew

to recalibrate the altimeter to a different base pressure,allowing for variations in atmospheric pressure atany airport.

• The current altimeter setting for a given airport is provided

to the pilots before takeoff, or during the approach, so thatthey may adjust the altimeter to indicate properly.

Altimeter Settings



A4, Page 41


The ability to shift the calibration tothe left or right allows the altimeter toshow the same altitude regardless of

the local air pressure.Indicated Altitude

29.92

H i g h - P

r e s s u r e D a y

S t a n d a

r d D a y

L o w - P r e s s u r e D

a y

Altimeter Setting



A4, Page 42

• There are three specific altimeter settings with which weshould all be familiar. They are referred to as:

QNH, QFE, and QNE

• “QNH” is the term used to designate the most frequentlyused altimeter correction. When QNH is set on an altimeter,

the altimeter will then read pressure height above sea levelfor the local prevailing atmospheric pressure conditions.Meaning, the altimeter will read the actual airport altitudewhen the airplane is sitting on the ground at the airport.

Obviously, QNH varies from day to day for a given location,and varies from location to location on a given day. It isthus of only local usefulness.

QNH is the altimeter setting



A4, Page 43


QNH is the altimeter setting

which will cause the altimeter toshow airport elevation when the

airplane is on the ground.

Indicated Altitude

29.92

H i g h - P

r e s s u r e D a y

S t a n d a r d D

a y


a y

QNH

S t a n d a r d D

a y

Below29.92

Airport Elevation

Above29.92

Altimeter Setting



A4, Page 44

“QFE” is the term used for another altimeter setting,

which is used less widely. When set to QFE, analtimeter will show pressure height above localground level, rather than above sea level. Meaning,the altimeter will read zero when the airplane is sitting

on the ground at the airport. QFE varies from day today, and from location to location, depending onatmospheric variations.

QFE is the altimeter setting



A4, Page 45


QFE is the altimeter setting

which will cause the altimeterto show zero altitude when the

airplane is on the ground.

Indicated Altitude

29.92

H i g h - P

r e s s u r e D a y

S t a n d a r d D

a y


a y

QNH

S t a n d a r d D

a y

Below29.92

Airport Elevation

Above29.92

L o w - P

r e s s u r e D a y

S t a n d

a r d D a y

QFE

H i g h - P

r e s s u r e D

a y

0

Altimeter Setting



A4, Page 46

• For long-distance navigation, especially in airspace wheremany airplanes are navigating simultaneously (high

altitudes), all airplanes should be using the same altimetersetting so that they will all be reading their altitudes on acomparable basis.

• QNH and QFE are both subject to local variations, andhence are unsuitable for navigation except in the local areaaround the departure or destination airport.

• At all times when flying above the “transition altitude,” the

altimeter setting called “QNE” is used. QNE is the standardday setting of 29.92 inches Hg, or 1013.25 mb.

QNHOn ground, altimeter



A4, Page 47

QFE

QNE

reads airport elevation

On ground, altimeterreads zero

altimeter readspressure altitude

Elevation 1000

Elevation 1000

Transition Altitude

1000

0000

33000

Altimeter Errors Due to Non-Std Temperatures



A4, Page 48

• Another real, and potentially dangerous, inaccuracyin altimeters is due to variation in air density.

• As we have said previously, all altimeters necessarilyassume some relationship between air pressure andheight. The ISA relationship is normally used forthis purpose.

• However, in atmospheric conditions which are colderor warmer than ISA, the relationship betweenpressure and actual height will not follow thestandard atmosphere. This is because the air density

will be different from the standard density when thetemperatures are above or below the standardtemperature. In those cases, the density comparedto the standard would be less or greater respectively.

(ISA)(Hot Day) (Cold Day)

Effect of Density on Pressure-Height Relationship



A4, Page 49

(ISA)(Hot Day) (Cold Day)

Density:1.0 kg/m3

Each box

is 1 m3

Pressure, kg/m2

10.0

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

∆p of 1 kg/m2

= ∆ h of 1 m

Density:0.9 kg/m3

9.0

8.1

7.2

6.3

5.4

4.5

3.6

2.7

1.8

0.9

∆p of 1 kg/m2

= ∆ h of 1.11 m

Density:1.1 kg/m3

11.0

9.9

8.8

7.7

6.6

5.5

4.4

3.3

2.2

1.1

∆p of 1 kg/m2

= ∆ h of .91

m




A4, Page 50

• Another way to think of this is that the slope of the calibrationcurve used by an altimeter to relate ∆pressure to ∆height isthe std day slope. For example, as air pressure decreasesfrom 29.92 to 28.86 in. Hg the altimeter will show an increase

in altitude of 1000 ft; the std day relationship.

• When air density is less than standard day (temperatureswarmer than std) then the air pressure decreases less rapidly

with increasing height. Thus, when the altimeter indicates theairplane has climbed 1000 ft, it has actually climbed more than1000 ft; the crew is higher than they think they are.

• Conversely, when air density is greater than standard day

(temperatures colder than std) then the air pressure decreasesmore rapidly with increasing height. Thus, when the altimeterindicates the airplane has climbed 1000 ft, it has actuallyclimbed less than 1000 ft; the crew is lower than they think

they are; potentially a problem if obstacles are present.

QNH and QFE only shift the altimeter calibration




A4, Page 51

∆p

QNH and QFE only shift the altimeter calibration

to the left or right (x-intercept), making it readcorrectly on the ground (airfield elevation for QNH

or zero for QFE) regardless of local

air pressure.

But, the slope of the calibration, i.e., indicatedheight versus change of pressure, does not vary.

∆ alt 3

∆ alt 1

∆ alt 2


Elevation

H i g h - D

e n s i t y D a y

S t a n d a r d D

a y

L o w - D

e n s i t y

D a

y

Air temperature affects air density: Cold air is more dense than warm air!




A4, Page 52

During departure or approach,

indicated change of height is

change of pressure height, notactual change of elevation.

Change of elevation, for a givenchange of pressure height,

depends on density of the air -

thus, on air temperature

∆ alt 1

∆

a l t 1

∆ alt 2

∆ alt 3

∆ Indicated Altitude

∆ Elevation

H o t

D a y

( L o w e

r D e n s

i t y )

S t a n

d a r d

D a y

C o l d D a

y ( H i g

h e r D

e n s i t y

)

Increasing

Increasing




A4, Page 53

• Remember, tapeline height and pressure height wouldonly be the same under actual standard day conditions.

Under any other conditions they will be different.

• Flight crews must never make the mistake of believingthat the altimeter displays tapeline height – physical

height above the terrain. Such an error couldconceivably, under certain conditions, cause them to flyinto a mountainside.

• On a cold day, the airplane’s true height above the

ground may be less than the crew believes (“cold andlow, look out below”)!



A4, Page 54

End of Aerodynamics (4):Standard Atmosphere/ Altitude Measurement

FLIGHT

OPERATIONS

ENGINEERING



A5, Page 1

Aerodynamics (5):

Air That is Moving / Airspeed Measurement

Aerodynamics (5): Air That Is Moving



A5, Page 2

• Streamlines and Stream tubes

• Continuity and Momentum Equations

• Bernoulli’s Equation

• Dynamic Pressure and Impact Pressure

• Airspeed Measurement

Streamlines

• A stream of air consists of countless particles moving in the



A5, Page 3

p g

same direction. The path traced out by any one particle in thatfluid stream, in a steady flow, is called a streamline.

• Streamlines assist us in visualizing the motion of a flowing gas,

and define the direction of the flow.• In steady flow, there is no flow across streamlines

Example 1:

Example 2:

E l e m e n t a

l

p a r t i c l e

o f f l u i d

Stream tube



A5, Page 4

i n c o m

i n g

f l o w

o u t g o i n g

f l o w

• A “stream tube” is a bundle of “stream lines” whichtogether form a closed curve or tube. There can be noflow through the walls of a stream tube because the wallsof the stream tube are streamlines.

• You can think of a stream tube as representing the waythe particles of the air or fluid are flowing at any giveninstant of time.

Conservation of Mass

• Mass of fluid flowing through a pipe or stream tube per unit



A5, Page 5

g g p p p

time in a steady flow is given by the equation:

flow

direction

V1 dt

V1, A1, ρ1

mass per unit time = m = ρ x V x A

V2, A2, ρ2

V2 dt

m = ρ1 x V1 x A1 = ρ2 x V2 x A2

• Mass of air passing one section of a pipe or stream tube

must be the same as the mass passing any other section inthe same amount of time. Therefore, the mass of air flowingby station 1 per unit time is the same as that flowing bystation 2 per unit time:

Continuity Equation



A5, Page 6

• Generalizing, we can say that for steady fluid flowthrough a pipe or stream tube, the mass flow per

unit time is constant:

• This is known as the Continuity Equation. It relatesdensity, velocity, and cross-sectional area at one

section of a stream tube to any other section of thestream tube, based on conservation of massflowing through the tube.

m = ρ x V x A = constant

Compressible Versus Incompressible Flow



A5, Page 7

• Compressible Flow: A flow in which the density ofthe fluid can change from point to point in the flow.

All fluids in real life are compressible to somedegree, i.e. their density will vary as the pressurevaries. Air is highly compressible, especially at highspeeds.

• Incompressible Flow: A flow in which the density ofthe fluid remains constant from point to point in theflow, regardless of a change in pressure. Liquids

flows can be accurately treated as incompressibleflows. Air flows less than about 200 knots (~ Mach.30) can be approximated very closely asincompressible flows.

Continuity Equation in Another Form



A5, Page 8

• We can take the differential of the Continuity Equation toproduce another useful form as follows:

• Assuming an incompressible fluid flow (therefore dρ = 0)we can simplify this equation to:

• A convergence in cross-sectional area of a stream tube willproduce an increase in velocity, whereas a divergence in

area will produce a decrease in velocity.

dρ VA + dV ρ A + dA ρV = 0

ρ x V x A = constant

dρ + dV + dA = 0

ρ V A

(Incompressible Flow)dV dA

V A= -

(Compressible Flow)

Momentum Equation



A5, Page 9

Let us consider an infinitesimally small fluid elementmoving along a streamline with velocity = V:

streamline

V

dy

dx

dz

p p + dp dx

dx

Looking at the forces acting on this fluid element (neglectingfriction and gravity):

F = ma

p + dx (dy dz) =

dx

dpp (dy dz) - ρ (dx dy dz)V

dx

dV

Momentum Equation



A5, Page 10

• Simplifying the equation by canceling similar terms leads to:

• This is known as the Momentum Equation (also known asEuler’s equation). It relates rate of change of momentum toforce; relating a change in pressure to a change in velocity.

• For compressible flow, ρ is a variable. For incompressibleflow, ρ is considered to be a constant.

• Let us use this equation to derive an equation you haveprobably all heard of, the Bernoulli equation. We will look firstat the incompressible form of the Bernoulli equation.

dp = - ρ V dV

Assuming incompressible flow (i e ρ is constant) and

Incompressible Bernoulli Equation



A5, Page 11

Assuming incompressible flow (i.e., ρ is constant), andintegrating the momentum equation between two pointsalong a streamline, point 1 and point 2:

dp = - ρ V dV

dp = - ρ V dV

p1

p2

∫V1

V2

∫p2 – p1 = - ρ V2 – V1

2

2 2

2

p2 + ρ =V2

2

2p1 + ρ

V1

2

2

This is what is known

as the ‘BernoulliEquation.’

It is presented in its

more general form on

the following page

Incompressible Bernoulli Equation

2



A5, Page 12

• Remember, this relationship is only good for incompressibleflows; i.e., airspeeds less than approximately 200 knots.

• In incompressible flow, static air pressure plus any amount of

pressure due to velocity is called ‘total pressure.’ Totalpressure remains constant along a streamline, and equalsthe pressure that would be measured if the flow was broughtto zero velocity.

• This equation can be used to relate pressure and velocityalong a streamline in a pipe or stream tube. You can see thatincreasing the flow velocity decreases the static pressure, astotal pressure remains constant.

total

2truestatic

pstreamlineaalongconstantV21p

==ρ+

Dynamic Pressure, “q”



A5, Page 13

• One term in the preceding equation which is encounteredfrequently in aerodynamics is given the name “dynamic pressure”and is denoted by the symbol ‘q’:

• Dynamic pressure, q, is a measure of the kinetic energy of the flow.

pressuredynamicqV2

1 2

true

==ρ

Airspeed Measurement



A5, Page 14

• Bernoulli’s incompressible flow equation states that staticpressure plus dynamic pressure is constant, and equal to totalpressure. Conversely, total pressure minus static pressure

equals dynamic pressure, which is a direct function of airspeed.

• If we had a device that could measure pt – ps, and we were onlyconcerned with relatively low-speed airflows, we could us thisequation to calibrate that device in units of airspeed. One suchdevice is called a manometer.

( )ρ

−=

ρ==−

sttrue

2truestatictotal

pp2V

V21qpp


A manometer is a U-shaped tube partially filled with a fluid



A5, Page 15

A manometer is a U shaped tube partially filled with a fluid.One side of the tube is connected to the total pressuresource, while the other side is connected to the staticpressure source.

ptotal pstatic

∆h

p2

p1

Applying the hydrostatic equation to the fluid on theright side of the tube:

p2 - p1 = ρfluid g ∆h

But, p2 = ptotal, and p1 = pstatic Therefore:

ptotal - pstatic = ρfluid g ∆h

Combining this with our velocity equation From theprevious page:

( )

air

fluidtrue

hg2V

ρ

∆ρ=




A5, Page 16

• Obviously, a manometer would not be very practicalfor measuring airspeed on an airplane.

• Instead, airplanes use a system called a pitot-staticsystem to obtain the measurements of total andstatic pressure used in the determination of

airspeed.




A5, Page 17

A pitot-static probe combines the total and static pressuremeasurements into one device.

Total pressurefelt here

Static pressurefelt hereVelocity

Mechanical/electricaldevice measures ∆p

∆p =pt - ps




A5, Page 18

Total pressurefelt here

Static pressurefelt here

Velocity

∆p =pt - ps

∆p is called ‘Impact Pressure’

Another version of a pitot-static system uses a pitottube combined with the static source located in aseparate location.

Impact Pressure Versus Dynamic Pressure



A5, Page 19

• Impact pressure and dynamic pressure are not the same!

• Only at low airspeeds, where the air flow behaves similarto an incompressible flow, does the incompressibleBernoulli equation adequately model that flow. In thatsituation, dynamic pressure will equal impact pressure.But remember that they are defined differently.

2true

statictotal

V21qPressureDynamic

ppPressureImpact

ρ==

−=

Pitot-Static Systems on Boeing Airplanes



A5, Page 20

Pitot-static probes apply to models:

• 737-200, -300, -400, -500

• 747-100, -200, -300, -400, SP

• 767-200, -300, -400

Flush static ports apply to models:

• 707-300

• 727-100, -200

• 737-600, -700, -800, -900

• 757-200, -300

• 777-200, -300




A5, Page 21

• How shall we calibrate this airspeed system, based on thesepitot-static measurements of impact pressure?

• If we always flew at low speeds, such that the air flow could beconsidered incompressible, we could use the incompressibleBernoulli equation to obtain airspeed:

• However, commercial aircraft fly at speeds where the effects ofcompressibility are large and must be considered.

• To accurately model high-speed, compressible flows, someadditional complexity must be added to the incompressible

Bernoulli equation.

( )ρ−= sttrue pp2V

Compressible Bernoulli Equation



A5, Page 22

• Real, high-speed flows are compressible.

• To correctly analyze them it helps to consider the concept of

conservation of energy in the flow through a stream tube (i.e.,energy is neither created or destroyed, but may be changed inform; total energy is constant).

• Conservation of energy states that the total energy of a flowingfluid is made up of both thermal energy and kinetic energy.For two locations along a stream tube this can be stated as:

T h e r m a l

e n e r g y

P e r u n i t

m a s s K i n e

t i c e n e r g

y

P e r u n i t

m a s s

totalp222

211p TCconstantV

21TCpV

21TC ==+=+

Compressible Bernoulli Equation

• We can convert the preceding ‘conservation of energy’



A5, Page 23

p g gyequation into a more “recognizable” form with the help ofsome definitions we covered earlier in the class:

T =

p

ρR R = Cp - Cv =

Cp

CvT =

p

ρ(Cp - Cv)

Cp T =p

ρ

( - 1)

• Applying this to the conservation of energy equation createsthe Compressible Bernoulli Equation:

+ V1

2

2

1= constant

p1

ρ1

( - 1)

pt

ρt

( - 1)=




A5, Page 24

• The compressible Bernoulli equation is the basis for all modernaircraft airspeed systems.

• Remember that the pitot-static system measures impact

pressure (pt – ps). A few further adjustments are necessary inorder to convert the compressible Bernoulli equation into a formmore useable in the determination of airspeed from ameasurement of pt – ps.

• For Isentropic Flow(no heat transfer and frictionless):

• Using the above, and recognizing that:

• We can rearrange Bernoulli’s equation to produce Velocity as afunction of : pt – ps, ps, and ρs

ptρt

1

= p

pt=p

pt - pp + 1




A5, Page 25

• The preceding equation is the basis for the calibration of allmodern aircraft airspeed instruments.

• The accuracy of an airspeed instrument depends on theaccuracy of the measurement of the impact pressure, and anumber of other factors.

• One of these other factors is that the pitot-static systemmeasures pt-ps, but what values should be used for the otherps and ρs terms in the calibration of the instrument?

• The following pages will address these concerns and lead us

through the definitions of: IAS, CAS, EAS, and TAS.

( )

( )

−

−

ρ−γ

γ =

γ −γ

1p

pp

1

p2V

1

s

st

s

sTAS

Indicated Airspeed



A5, Page 26

• Airspeed indicators are calibrated by using sea level, standardday values for ps and ρs in the calibration of the instrument.

• As a result of this, the airspeed indicator displays what is calledIndicated Airspeed.

• IAS would only equal TAS on a sea level, standard day(assuming no errors in the measurement of pt – ps).

(po , and ρo are sea level, standard day values, γ = 1.4)

( )

( )

−

+

−

ρ−γ

γ =

γ

−γ

οοο 11p

pp

1

p2V

1

stIAS

Calibrated Airspeed



A5, Page 27

• Calibrated airspeed is the result of correcting indicated airspeedfor errors in the measurement of static pressure (in the termpt – ps).

• The majority of this error is due to the location of the staticpressure source. The measurement of static pressure is greatlyinfluenced by the flow field around the airplane, especially asthe airplane is operated through a large range of angles of

attack, such that the pressure sensed at the static port may bedifferent from the free-stream static pressure.

• During flight test, a ‘trailing cone’ is used to measure the truefreestream static pressure for comparison to the aircraft

mounted static source. The results are used to determine thecalibration correction.

• The calibration correction is referred to as “position error,” or,“static source error” correction (SSEC), and denoted as ∆Vp.

Calibrated Airspeed

Trailing



A5, Page 28

Cone

MN 29901 = 5’4’

3’2”1’



A5, Page 29

Cable (Attaches cone to stinger)

Trailing cone

Trailing cone AirplaneStinger

Static ports (Holes)

Radiator hose hides1’ of hose from view

Fin

Stinger Radiatorhose

Tail conehoseStatic ports are located onthe stinger, 10” behind theconnection to the hose

Calibrated Airspeed

Example of Static Source Error Correction



A5, Page 30

1 2 0 1 1

0 1 0 0

9 0 8 0

7 0 6 0 G R O

S S W E I G

H T ~ 1 0 0 0

K G

140 160 180 200 220 240INDICATED AIRSPEED ~ KNOTS

-6

-4

-2

0

2

V p ~ K N O T S

140 160 180 200 220

INDICATED AIRSPEED ~ KNOTS

1 2 0 1 1

0 1 0 0 9 0

8 0 7 0

6 0 G R O S S

W E I G H

T ~ 1 0 0 0

K G

FLAPS 0, 1 FLAPS 5

-150

-100

-50

0

50

H p ~

F E E T

-150

-100

-50

0

50

H p

~ F E E T

1 2 0 1 1 0

1 0 0 9 0

8 0 7 0

6 0 G R O S

S W E I G H T

~ 1 0 0 0 K G

1 2 0

1 1 0 1 0 0

9 0 8 0 7 0 6 0

G R O S S W E

I G H T ~ 1 0 0

0 K G

-8

-6

-4

-2

0

2

V p ~ K N O

T S

Charts are provided in the AFM, or the correction is builtinto the airspeed system

Calibrated Airspeed

• After applying the position error correction to the indicated airspeed,



A5, Page 31

the resulting speed is referred to as the calibrated airspeed.

• ∆Vp is different for each airplane model, and sometimes for differentengines on the same model.

• Looking again at the airspeed equation:

(po, and ρo are still sea level, standard day values, but pt - ps

has been corrected for errors in its measurement)

pIASCAS VVV ∆+=

( )

( )

−

+

−

ρο−γ

γ =

γ

−γ

11

pο

pspt

1

pο2VCAS

1

Equivalent Airspeed

If VCAS i dj t d f th t t ti t th i



A5, Page 32

• If VCAS is adjusted for the correct static pressure at the given pressurealtitude, we obtain what is called equivalent airspeed, VEAS.

• This adjustment has been put into terms of airspeed in the chart onthe following page, such that:

• ∆VC is referred to as the ‘compressibility correction.’ It should not beconfused with going from incompressible to compressibleflow equations.

(ρo is still the sea level, standard day value)

CCASEAS VVV ∆−=

( )

( )

−

+

−

ρ−γ

γ

=

γ

−γ

ο 11p

pp

1

p2

V

1

s

sTs

EAS



A5, Page 33 A5, Page 33

True Airspeed

• Finally, if the correct value of ρ is used in the equation, we obtaini d V b d di li i f B lli’



A5, Page 34

strue airspeed, VTAS, based on a direct application of Bernoulli’scompressible flow equation.

• VTAS can be obtained from VEAS by use of the following:

( )

( )

−

+−

ρ−γ

γ =

γ

−γ

11p

pp

1

p2V

1

s

sT

s

sTAS

σ= EASTAS

VV

Remember: θδ

=ρο

ρ

=σ

= EASV θδ

S d di l d th i d i di t

Summary of Airspeed Relationships

IASV



A5, Page 35

Speed displayed on the airspeed indicator

Adjusted for errors in measurement of ps

Adjusted for correct ps at given pressure altitude

Adjusted for correct ρs

IASV

CASV

EASV

TASV

pIASCAS VVV ∆+=

CCASEAS VVV ∆−=

σ= EAS

TASVV

Equivalent Airspeed and ‘q’

VEAS

seems to be just an intermediate step between VCAS

andV Wh t j t t ti t di tl f V



A5, Page 36

EAS CASVTAS. Why not just create a correction to go directly from VCAS

to VTAS and skip VEAS?

1. Temperature is needed in order to calculate VTAS

.

2. VEAS has a unique relationship to dynamic pressure.

2TASV

21

ο σρ=

2TASV21q ρ=

A constant equivalentairspeed produces a

constant dynamicpressure!

EASV

21q ορ= 2

V

σ=

σ

=

σ=

2TAS

2EAS

2EAS2

TAS

EASTAS

VV

VV

V

Other Airspeed Relationships



A5, Page 37

• You have seen the steps required for obtaining trueairspeed when given indicated airspeed, going through a

series of corrections for different errors.

• There exist a number of equations relating various formsof airspeed: for example, finding true airspeed when

given calibrated airspeed, and so on.

• The “Summary of Useful Equations for PerformanceEngineers” contains a complete list of these equations,

repeated on the next few pages for convenience.

CAS to EAS, or, EAS to CAS

Requires Knowing δ



A5, Page 38

( )

( )

−

+

−

δ+δ=

−

+

−

+δδ=

1111.1479

EAS1

11.1479CAS

1114786.661CAS

2.11

1.1479EAS

5.31

5.32

5.31

5.32

CAS to TAS, or, TAS to CAS

Requires Knowing δ and θ



A5, Page 39

( )

( )

−

+

−

θ+δ=

−

+

−

+δθ=

1111.1479

TAS1

11.1479CAS

1114786.661CAS

2.11

1.1479TAS

5.31

5.32

5.31

5.32

EAS to TAS, or, TAS to EAS

Requires knowing σ, or δ and θ



A5, Page 40

δθ=

σ= EASEASTAS

θδ

=σ= TASTASEAS



A5, Page 41


Air That Is Moving

FLIGHT

OPERATIONS

ENGINEERING



A6, Page 1

Aerodynamics (6):

Air That Is Moving Nearthe Speed of Sound

Aerodynamics (6): Air That Is Moving Near the Speed of Sound



A6, Page 2

• Sound and the Speed of Sound

• Mach Number

• Flow Relations Near the Speed of Sound

• Shock Waves

• Total Temperature, Pressure, and Density



A6, Page 3

Video: “Approaching the Speed of Sound”

(Part I of the Shell Oil Company series,

“High Speed Flight”)

Sound

• “Sound” is simply our perception of a series of “waves” orpulses of air pressure as sensed by the eardrum



A6, Page 4

pulses of air pressure as sensed by the eardrum.

• A pressure “wave” is characterized by a compression(pressure increase) at the front of the wave and a rarefaction

(pressure decrease) behind the wave.

Pressure WavesPressure Waves

Speed of Sound



A6, Page 5

• Just as waves in water travel at a predictable speed, so dowaves of pressure in a gas.

• The rate at which a pressure wave travels in a gas dependson how easily that gas can be compressed. The moreeasily a gas may be compressed, the slower the velocity will

be of a pressure wave traveling in that gas.

• It can be shown that the speed of sound in a gas, called ‘a’,follows the equation:

TRa ××γ =

Speed of Sound



A6, Page 6

• The speed of sound at sea level standard day temperature,15 deg. C, is 661.4786 knots. This is referred to as ‘ao’.

• It can be shown that the speed of sound in a gas for anytemperature may be expressed as:

• The speed of sound depends only on temperature!

θ∗=θ∗ο= 4786.661aa

Mach Number



A6, Page 7

• “Mach number” is defined as an object’s true airspeed expressedas a decimal fraction of the speed of sound at the ambient

conditions surrounding the object.• For example, when flying at the speed of sound, the Mach

number is 1.0, when flying at one half the speed of sound theMach number is 0.5.

• In terms of airspeed in knots, Mach number can be expressed as:

θ∗=Μ

4786.661)knots( AirspeedTrue =

aVTAS



A6, Page 8

Question #1:

For an OAT of –50 degrees C, and a true airspeedof 450 knots, what is the Mach number?



A6, Page 9

Question #2:

For the following conditions calculate: VTAS, VEAS, VCAS

Flight Level = 330, ISA + 10 deg. C, M = .80

Flow Relations Near the Speed of Sound

A CB



A6, Page 10

Lower sourcepressure

Increasing sourcepressure

“Critical” sourcepressure

speedsubsonic

speedIncreases

(still subsonic)

sonic

speed

speedincreases

static pressuredecreases

speeddecreases

static pressureincreases

speedincreases

static pressuredecreases

PressureSource

AConverging

B

“Throat”

CDiverging

(same)

(same)

A CB


In a converging-diverging nozzle connected to a source of gasunder pressure the following relations hold true:



A6, Page 11

In a converging diverging nozzle connected to a source of gasunder pressure, the following relations hold true:

• At lower source pressures:

– In the converging section, velocity will increase and staticpressure, temperature, and density will decrease.

– Velocity will be the highest at the smallest cross-section(called the “throat”).

– In the diverging section, velocity will decrease and staticpressure, temperature, and density will increase.

• As source pressure increases:

– Velocity at the throat will increase.

• At some value of source pressure:

– Velocity at the throat will reach the speed of sound.

It b h th t f ibl fl id ( ll fl id i lit




A6, Page 12

• It can be shown that for compressible fluids (all fluids in reality,but air must be treated this way at speeds above approximately200 knots) the relationship between velocity and cross section is

a function of Mach:

• Thus, when the flow velocity reaches the speed of sound at thethroat, the diverging section will cause a further increase invelocity and a decrease in pressure, temperature and density.

(Area – Velocity Relation) (Area – Pressure Relation)

( )V

dV1d 2 −Μ=ΑΑ

Ρ Ρ

Μγ

Μ−=ΑΑ d1d

2

2

Subsonic nozzle

M < 1M < 1

M < 1

For M < 1 (subsonic flow), the

rates of change of V and A haveit i Thi i th




A6, Page 13

Supersonic nozzle

Supersonic Diffuser

M 1A V P M < 1

M 1A V P

M < 1

A V P M = 1

M > 1

A V P

M > 1

A V P M = 1

M < 1

A V P

rates of change of V and A haveopposite signs. This is the same

as in incompressible flow.

For M > 1, the rates of change of

V and A have the same sign. This

is the opposite of incompressible

flow and subsonic flow.

For M = 1 (sonic flow), the rate of

change of V can only be finite

when the rate of change of A =0.

In other words, Mach = 1 will only

occur at a local max. or min. in

area. For a convergent–divergent

duct this occurs at the throat.

• At some point in a stream tube, if the free-stream (ordownstream) flow is subsonic then the supersonic flow will have




A6, Page 14

downstream) flow is subsonic, then the supersonic flow will haveto decelerate to subsonic velocities.

• This deceleration from supersonic to subsonic flow usually

happens through a “shock wave.” Downstream of the shockwave, the static pressure, temperature and density all increase.

• A shock wave is a drastic transition from supersonic to subsonic

flow, creating a virtually instantaneous change in V, P, ρ, and T.

Shock wave

M < 1M > 1

>V2

< P2

< ρ2

< T2

V1

P1

ρ1

T1

Effects of Compressibility onTemperature, Pressure and Density

Ab d f i t l 200 k t



A6, Page 15

• Above speeds of approximately 200 knots, we can nolonger simplify our calculations by assuming air acts asan incompressible fluid. At low speeds that assumption isacceptably accurate.

• When a fast-moving flow is brought to a standstill, it willbe compressed to some degree. The faster the flow, thegreater the compression. An example would be the flow

at the leading edge of the wing. At some point, calledthe “stagnation point,” the airflow is completely haltedrelative to the wing, and thereby highly compressedrelative to its free-stream, static condition.

• Compression increases the pressure and temperature ofthe air. The amount of increase depends on the speed ofthe flow.

• The temperature pressure and density of the air in its




A6, Page 16

• The temperature, pressure, and density of the air in itscompressed state are referred to as “total temperature,” “totalpressure,” and “total density.”

• Total temperature can be computed from the following equation:

• Remembering that γ = 1.4 for air, the above equation becomes:

(Temperature must be in absolute temperature; i.e., oR or oK)

Μ×

−γ +∗Τ=Τ 2

statictotal 2

11

( )2

2.1 Μ+∗S

ΑΤ=ΤΑΤ




A6, Page 17

It can similarly be derived that:

and,

( ) 5.32statictotal

2.1 Μ+∗Ρ =Ρ

( ) 5.22statictotal

2.1 Μ+∗ρ=ρ




A6, Page 18

θtotal = θstatic x 1 + .2 M

2

δtotal = δstatic x 1 + .2 M

23.5

σtotal = σstaticx 1 + .2 M

22.5

Total Temperature Ratio

Total Pressure Ratio

Total Density Ratio

In airplane performance calculations we tend to work morewith total temperature, pressure and density ratios:



A6, Page 19

Question #3:

OAT = –50 deg. C pressure altitude = 33,000 feet

Mach = 0.8What are the values of: θtotal, δtotal, σtotal?



A6, Page 20

End of Aerodynamics (6): Air That Is Moving Near the

Speed of Sound

FLIGHT

OPERATIONS

ENGINEERING



A7, Page 1

Aerodynamics (7):

Flow Near the Surfaceof an Object

Aerodynamics (7):Flow Near the Surface of an Object



A7, Page 2

• More on Viscosity

• Non-Viscous versus Viscous Flow

• The Boundary Layer

• Reynolds Number

Viscosity

• Viscosity may be thought of as the "stickiness“ of a fluid.

– Viscosity tends to slow down the movement of an object



A7, Page 3

Viscosity tends to slow down the movement of an objectthrough the fluid.

– E.g., Water versus oil; oil is more "viscous" than water.

• Viscosity is a measure of the internal friction within a fluidcaused by molecular forces exerted as one molecule movesrelative to another molecule, or, a measure of a fluid’s

resistance to shear when in motion.• The shear stress within a fluid is proportional to the coefficient

of viscosity, µ, times the transverse gradient of the velocity.

τ = µdv

dysurface

yv

Viscosity of Air



A7, Page 4

• Air has viscosity too, although it’s of course much less thanwater or oil.

• Coefficient of viscosity for air is assumed to vary withabsolute temperature according to ‘Sutherland’s’ equation:

T is temperature in degrees Kelvin

0.3125059 x 10-7 x T1.5

T + 120

(LB)(sec)

FT2

Question #1a:



A7, Page 5

For air at a temperature of 15 deg. C, what is the

value of µ?

Question #1b:

For air at a temperature of 40 deg. C, what is thevalue of µ?

Fluid Flow Over a Flat Plate



A7, Page 6

In a non-viscous flow, the velocity of the fluid flowing over asurface would be uniform, regardless of the distance from the

surface. There would be no transverse gradient of velocity.

Flowdirection

Vo

Vο = free stream velocity

yNon-viscous flow

Uniform velocity distribution

Fluid Flow Over a Flat Plate

• In viscous flow, the velocity of the flow right at the surface

is reduced to zero by viscosity.



A7, Page 7

Flowdirection

Vo

Vο

= local free stream velocity

y

Viscous flow causes

Non-uniform velocity distribution

Boundarylayer

• The tendency of one layer of fluid to drag along the layernext to it, i.e., viscosity, slows the fluid flow as we get closer

to the surface.

• There is a finite thickness of fluid next to the surface whichis slowed relative to the free-stream velocity. This layer is

called the “boundary layer.”

Boundary Layer



A7, Page 8

• The finite thickness of fluid next to a surface which isslowed relative to the "free stream" velocity of the flow

is called the boundary layer.

• The thickness of the boundary layer depends on:

– Viscosity of the fluid

– Distance along the surface

– Nature of the flow in the boundary layer

Viscosity Effects on Fluid Flow Over a Surface

V



A7, Page 9

• The flow immediately downstream of the leading edge issmooth and is known as laminar flow. Further downstream, thethickness of the layer increases as more and more air next to

the surface becomes affected by viscosity.• As the flow progresses downstream along the surface, a point

is reached where the laminar flow breaks down and becomesunsteady. This is called the transition region, where theboundary layer thickness increases rather suddenly.

Laminar

Transition Region

Turbulent

L

V

V

Viscosity Effects on Fluid Flow Over a Surface



A7, Page 10

Laminar

Transition Region

Turbulent

L

V

• Beyond the transition region, the air becomes even moreturbulent due to considerable fluid particle motion. This regionis called the turbulent boundary layer region.

• The thickness of the boundary layer, in both the laminar andthe turbulent regions, and the transition location can bepredicted and expressed in terms of a parameter called theReynolds Number, RN.

Reynolds Number

• Reynolds number, a very important parameter in viscousflow, is defined as:



A7, Page 11

• RN is a parameter which indicates the ratio of inertial forces

to viscous forces.

• Under ideal conditions, the transition from laminar toturbulent flow occurs at a RN of approximately 530,000.

RN =µ

(Dimensionless)

where: ρ is air density

V is free-stream velocity

L is distance along surfaceµ is viscosity of the fluid

ρVL



A7, Page 12

Question #2:

For a flow of air at sea level and 15 deg. C overan airfoil, at a freestream velocity of 200 feet per

second, how far back from the leading edgedoes the transition from laminar to turbulent flowoccur? (assume ideal conditions)

Reynolds Number



A7, Page 13

• From the previous question you have seen that thetransition from laminar to turbulent flow on the wing of a

typical jet airplane occurs very close to the leading edgeof the wing. Thus, the entire wing is usually consideredto have a turbulent boundary layer.

• Since Reynolds number affects the thickness of theboundary layer, it also affects the aerodynamic drag.Components of drag will be discussed later in the class.



A7, Page 14

Question #3:

If I want to test a small scale model of an airplane in awind tunnel, at the same conditions as in Question #2,

is it true that most or all of the model's wing would be inlaminar flow? If that is true, then would the aerodynamicdrag of the model truly represent (allowing for scale, ofcourse) the drag of the real full-size airplane?



A7, Page 15

End of Aerodynamics (7):Flow Near the Surface of

an Object

FLIGHT

OPERATIONS

ENGINEERING



A8, Page 1

Aerodynamics (8):

Viscosity Effects onPressure Distribution

Aerodynamics (8):Viscosity Effects on Pressure Distribution



A8, Page 2

• Ideal versus Viscous Flow Around a Cylinder • Ideal versus Viscous Flow Around an Airfoil



A8, Page 3

An understanding of the Bernoulli principleis fundamental to the understanding of

aerodynamic forces.

(static pressure and velocity move in

opposite directions)

Flow Over a Cylinder in Non-Viscousand Incompressible Fluid Flow

• The stagnation point is the point on the cylinder at which



A8, Page 4

g ythe flow is “normal” to the surface. This is the point atwhich some of the flow comes to rest before turning

around the cylinder.• The flow velocities are at a minimum at the front and rear

of the cylinder; they are at a maximum at the sides, at 90degrees to the free-stream flow, where the streamlinesare compressed the greatest amount.

• In accordance with the Bernoulli principle, the staticpressures are the greatest at the front and rear of the

cylinder, where velocities are the least; static pressuresare the least at the sides, where the velocity isthe greatest.

Flow Over a Cylinder in Non-Viscousand Incompressible Fluid Flow



A8, Page 5

Vo

Po

In this ideal fluid, static pressures are distributed symmetricallyover the cylinder. As a result, there is no net force on the cylinder –

no drag, no lift.

Flow Over the Same Cylinder with Viscosity



A8, Page 6

• Due to the viscosity, the velocity distribution and hence thepressure distribution will be different.

• The net force, in viscous flow, is called drag.

Vo

Po

Flow of an Ideal Fluid Over a Symmetrical Airfoil



A8, Page 7

Vo

Po

In an ideal fluid flow, again the pressures are symmetric,and the net force is zero.

Flow Over a Symmetric Airfoil In a Viscous Fluid



A8, Page 8

Vo

Po

In the case of the symmetric airfoil at no angle to the flow,the net normal force is zero, but there is a net longitudinal

force – the drag of the airfoil.



A8, Page 9

End of Aerodynamics (8):Viscosity Effects on

Pressure Distribution

FLIGHT

OPERATIONS

ENGINEERING



A9, Page 1

Aerodynamics (9): Airfoils

Aerodynamics (9): Airfoils



A9, Page 2

• Properties of Two-Dimensional Airfoils

• Lift, Drag, and Moment Coefficients• The Lift Curve

• The Drag Polar

Cross-Section of a Typical Airfoil

Maximum

Camber Mean Line



A9, Page 3

• Mean line: A line equidistant between the upper andlower surfaces.

• Chord line: A straight line joining the intersections of the meanline with the leading and trailing edges of the airfoil.

• Camber: The degree of curvature of the mean line.• Upper and lower ordinates: The distance of the upper and

lower surfaces from the chord line, usually expressed in inpercent of chord length.

Chord Line

Maximum Camber Mean Line

Chord Line



A9, Page 4

0 ----- 0 01.25 2.67 1.25 -1.232.5 3.61 2.5 -1.71

5.0 4.91 5.0 -2.267.5 5.80 7.5 -2.6110.0 6.43 10.0 -2.9215.0 7.19 15.0 -3.5020.0 7.50 20.0 -3.9725.0 7.60 25.0 -4.2830.0 7.55 30.0 -4.46

40.0 7.14 40.0 -4.4850.0 6.41 50.0 -4.1760.0 5.47 60.0 -3.6770.0 4.36 70.0 -3.0080.0 3.08 80.0 -2.1690.0 1.68 90.0 -1.2395.0 0.92 95.0 -0.70

100.0 ----- 100.0 0

NACA 23012 AIRFOIL(Stations and ordinates given in percent of airfoil chord)

UPPER SURFACE LOWER SURFACE

station ordinate station ordinate

• Consider an airfoil in a fluid stream (air), with the airfoil placedat an angle relative to the direction of the flow.

Air Flow Over a 2-Dimensional Airfoil



A9, Page 5

• The “angle of attack,” denoted by the Greek letter α (alpha) isthe term used to designate the angle of the chord line of the

airfoil relative to the remote direction of the air flow (the free-stream velocity vector)

• Note that angle of attack is not defined relative to a local orearth-based axis system.

Voα

C ho r d

Same angle of attack as airfoil on page 5.




A9, Page 6

V o

α

C h o r

d


Same angle of attack as airfoil on page 5.



A9, Page 7

Vo

α C h o r d


• Observe the stream lines in the photo on the following



A9, Page 8

Obse e t e st ea es t e p oto o t e o o gpage. Notice that the airflow over the top of the airfoil isdeflected more than the flow over the bottom, hence the

velocity over the top will be greater than over the bottom.• Notice also that the farther away from the airfoil, the less

the deflection of the streamlines. Moving far enough awaythe streamlines would be parallel to the free-stream airflow.

• This situation is similar to (but not exactly the same) as astream-tube having a reduction of cross section where theairfoil is located. Flow velocity above and below the airfoilis greater than the velocity ahead or behind the airfoil. Thisresults in a reduction of the static pressures above andbelow the airfoil.




A9, Page 9

(Actual photo from wind tunnel test. The lines are produced

by injecting smoke into the tunnel. Smoke trails follow thestreamlines of the air flow over the airfoil.)

• Similar to the symmetric airfoil seen earlier at no angle to the

flow, the flow velocities and resulting pressure distribution maybe shown for the airfoil at a positive angle of attack.




A9, Page 10

• The length of the arrows indicates the amount of pressurereduction or increase from the free-stream ambient pressure.

• Arrows are drawn perpendicular to the airfoil because pressurealways acts at a right angle to the surface.

Vo α

• The pressure forces may be resolved into a single equivalent

force. The location of that force on the wing is called the centerof pressure.




A9, Page 11

• There is no moment about the center of pressure, since the forcevector passes right through it and there is thus no moment arm

about the center of pressure.

Vo α

Resultant

C.P.

Lift and Drag Components

• The resultant force can be resolved into two components:

– Lift is the component of force normal to the direction ofremote air flow.



A9, Page 12

– Drag is the component of force parallel to the direction ofremote air flow.

Vo α

Resultant

C.P.Drag

Lift

Lift, Drag, and Moment about ¼ Chord

• For convenience, it is customary to resolve the lift and

drag forces about the quarter-chord point on the airfoil.• When relocating the lift and drag vectors to the quarter



A9, Page 13

Vo

C.P. Drag

Lift

VoDrag

Lift

s

c4

C.P.

Moment

When relocating the lift and drag vectors to the quarterchord, a moment must also be introduced in order tocreate forces and moments equivalent to original forcesacting at the C.P.

Aerodynamic Forces

Th f t d b i fl d d



A9, Page 14

• The forces generated by viscous flow depend onseveral variables:

– Flow velocity

– Surface area of the object

– Shape of the object

– Fluid density

– Viscosity of the fluid

• It is convenient to represent aerodynamic forces bysingle numbers which are independent of the flowvelocity and object size (area). This is possible throughthe use of force coefficients.

• Forces may be simplified to the form of a force coefficient:F

Force and Moment Coefficients



A9, Page 15

• Where: CF is a dimensionless coefficient of force

q is the dynamic pressure

A is a representative area, such as wing area

• Moments may also be expressed in a similar form:

• Where: L is a representative length to make thecoefficient dimensionless

Α=

×q

F C F

L AqC

××

Μ=Μ


• The forces and moment on the airfoil may be reduced to coefficientform using the force equations described previously, but now madespecific to the lift, drag, and moment of a wing.



A9, Page 16

p , g, g

where: q is the dynamic pressureS is a representative area (usually the wing area)

L is a representative length (usually the wing chord)

• These equations for CL, CD, and CM, may be expressed in anumber of different forms.

SqforceliftCL ×

=Sq

forcedragCD ×= LSq

momentCM ××=


where: ρ is air density

S is the ref wing areaVtrue is the true airspeed

SV2/1

LiftC 2

trueL ρ=



A9, Page 17

0023772841

CL

SV2/1

Lift2trueοσρ

=

CLSV

Lift4.8412true

σ=

.4. =

V in feet/sec:

V in knots:

Mach number:

CLSV

Lift369.2952true

σ=

SV

Lift369.2952equiv

CL =

CLS4.1481Lift 2δΜ=

×=

26878.1002377.

2369.295

5.0trueequiv VV σ=

= 369.2954786.6614.1481

2

S in units of FT2

Lift in units of LB


CD SV2/1

Drag2trueρ= CM LSV2/1

Moment2true ×ρ

=S in units of FT2

Lift in units of LBM in units of LB-FT



A9, Page 18

V in feet/sec:

V in knots:

Mach number:

CM =

LSV2/1

Moment2

true

×σρο

CMLSV

Moment4.8412true

×σ=

CDSV

Drag369.2952true

σ=

CSVDrag369.295

2equiv

D =

CD SM4.1481

Drag2δ= CM LSM4.1481

Moment2 ×δ=

CMLSV

Moment369.2952true

×σ=

CMLSV

Moment369.2952equiv

×=

CD

SV2/1

Drag2

trueοσρ

=

CDSV

Drag4.8412true

σ=

Q ti #1



A9, Page 19

Question #1:

If an airfoil having an area of 10 square feetgenerates 475.4 pounds of lift in a wind tunnel,

at a true airspeed of 200 ft/sec and a sea levelstandard day air density of 0.002377 slugs percubic foot, what is the value of CL?

Q estion #2



A9, Page 20

Question #2:

If the chord of that same airfoil is 2 feet, and thepitching moment measured about the quarter

chord is 95.08 ft-lb, what is the value of CM?

Question #3:



A9, Page 21

Question #3:

An airplane is flying in steady-state cruise (thatis, thrust and drag are equal) at FL330. The total

engine thrust is 12,000 pounds. The cruisespeed is Mach 0.80, and the wing area is 1951sq. ft. What is the value of CD?

The forces on the airfoil, and thus the force coefficients, will varyas the angle of attack varies. Below, is an illustration of thechange of lift coefficient with angle of attack.

Lift as a Function of Angle of Attack — The Lift Curve



A9, Page 22

flowseparation

begins

angle of zero lift

LiftCoefficient

CL

increasedseparation

CL max

angle of

maximumlift

Airflow in astalled condition:

Angle of attack (α)

(-) 0 (+)

Lift as a Function of Angle of Attack — The Lift Curve

• CL increases with increasing angles of attack.

A th l f tt k i th i i f d t k l



A9, Page 23

• As the angle of attack increases, the air is forced to make largerand larger changes of flow angle.

• At some angle, the flow will begin to separate from the wing,causing unsteady, turbulent airflow. This turbulence flows off thewing and hits the tail, causing buffeting of the airplane. The

speed at which buffet begins is called the initial buffet speed.• At a large enough angle of attack the airflow begins to break

down. This results in a loss of lift with further increases in angleof attack. This is a condition known as stalling the airfoil. Theangle of attack at which lift begins to decrease with furtherincreases in angle of attack is called the stall angle of attack.

• The drag coefficient may also be plotted versus angle

of attack.

D

Drag as a Function of Angle of Attack



A9, Page 24

• However, for airplane performance work it is generally moreconvenient to plot CL versus CD. This characteristic shape is

called the “drag polar.”

Angle of attack (α)

(-) 0 (+)

DragCoefficient

CD

The Drag Polar — Drag Versus Lift

Lift

CoefficientCL



A9, Page 25

• If a line is drawn from the origin to any point along the dragpolar, the slope of that line is a measure of the lift-to-dragratio; an important parameter in airplane performance.

Drag Coefficient, CD

CL

CDMAX

• The slope of a line drawn from the origin to a point just tangent

to the drag polar indicates the maximum lift-to-drag ratio.

Lift

CoefficientCL

The Drag Polar — Drag Versus Lift



A9, Page 26

Drag Coefficient, CD

CL

CDMAX

CL for (L/D) MAX

• Knowing the CL for maximum L/D allows for the calculation ofthe airplane speed which would offer the best lift-to-drag.

• Flying at the CL for maximum L/D allows the airplane to fly atits minimum angle of descent, or best angle of climb; allows

for optimization of the airplane’s performance.



A9, Page 27


Airfoils

FLIGHT

OPERATIONS

ENGINEERING



A10, Page 1

Aerodynamics (10): Wings

Aerodynamics (10): Wings

• Wing Terminology



A10, Page 2

Wing Terminology

• Planform Effects• 3-Dimensional Effects

• Wing Sweep and High Speed Airflow

• Drag Analysis

Three-Dimensional Wing Terminology

• Up to this point we have been discussing two-



A10, Page 3

Up to this point we have been discussing twodimensional airfoils, such as you might see in a

wind tunnel. Real airplanes, of course, have three-dimensional wings, and that fact introduces someadditional factors we must consider.

• We have already defined chord, camber, and so on. Additional terminology needs to be introduced fordiscussing three-dimensional wings.




A10, Page 4

Wingspan: The tip-to-tip dimension of the airplanewing. The symbol used for wingspan is the letter ‘b’.

b

CR




A10, Page 5

Taper Ratio: The ratio of the chord of the wing atits tip divided by its chord at the root. It is denoted

by the lower-case Greek character lambda, λ.

CT

RC

CΤ=λ


Λ, Sweepback Angle



A10, Page 6

• Quarter Chord: A line drawn along the span of thewing one-fourth of the chord behind the leading edge.

• Wing Sweep: The angle between a line perpendicularto the plane of symmetry of the airplane and thequarter chord of each airfoil section. The sweep angleis denoted by the upper-case Greek letter lambda, Λ.

C4


CR



A10, Page 7

• Reference Wing Area: Denoted by the letter S, or SREF,thereference wing area can be defined in a few different ways:

– Trapezoidal area (does not account for yehudies, gloves,strakes, etc.)

– ‘Effective’ area (Most commonly used definition at Boeingfor aerodynamic and performance work); accounts for the

complete projected outline of the wing.

CT




A10, Page 8

Chord of a rectangular wing(S = c * b)

Aspect Ratio (AR): This is defined as b2/S, and is ameasure of the ‘fineness’ of the wing planform. AR isequivalent to b/c for a rectangular wing. We will see later its

importance to airplane drag.

b


• Mean Aerodynamic Chord (MAC): The theoretical chord ofan untapered, unswept (rectangular) wing having the same

i f th t l i d th d i



A10, Page 9

wing area of the actual wing, and the same aerodynamicforces and characteristics throughout the flight regime.

• Used somewhat in airplane performance, but its mainconvenience is in the area of longitudinal S&C, and in denotingc.g. location in weight and balance.

• In Theory, the MAC could be determined by rigorous testing fora given wing. It depends on many factors including three-dimensional effects such as wing pressure distribution,

planform effects, and so on. However, it is customary andacceptably accurate to consider the MAC to be equivalent tothe mean geometric chord. The error in doing so is negligible.

MAC(Theoretical

0% MACLEMAC




A10, Page 10

MAC(Theoretical

rectangular wing)

100% MAC

The MAC passes through the centroid of the trapezoidal wing.

For a trapezoidal wing:

For any shape of wing:REF

2b

sdycMAC ο∫=

+λ +λ+λ= 1 1C32MAC

2

R

• For a wing of infinite aspect ratio (two-dimensional wing) the liftdistribution in the spanwise direction is constant.

Wing Lift Distribution: Two-Dimensional WingCompared to a Three-Dimensional Wing



A10, Page 11

Infinite aspect ratio wing (two-dimensional wing)

Finite aspect ratio wing (three-dimensional wing)

• For a wing of finite aspect ratio, the lift distribution varies overthe span of the wing.

rectangular elliptical tapered swept

PLANFORM

Shape of the Wing Planform Affects the Lift Distribution



A10, Page 12

1.0 1.0 1.0 1.0

LIFT DISTRIBUTION

SECTION LIFT COEFFICIENT DISTRIBUTION

Cl

CL

rectangular elliptical tapered swept

Shape of the Wing Planform Affects the Stall Characteristics



A10, Page 13

α << stall

α < stall

α = stall

• Rectangular Wing: The wing has good stallingcharacteristics, but poor high-speed performance. Non-elliptical lift distribution creates higher induced drag than

Shape of the Wing Planform Affects the Stall Characteristics



A10, Page 14

elliptical lift distribution creates higher induced drag thanan elliptical wing.

• Elliptical Wing: The entire wing tends to stall at once.Theory can show that the elliptical distribution of lift resultsin the minimum induced drag.

• Tapered Wing: Poor stalling characteristics because the tipstalls first. Lower induced drag than the rectangular wingbecause the lift distribution is more elliptical.

• Swept Wing: Poor stalling characteristics because the tipstalls first. High-speed drag benefit offering improvedaerodynamic efficiency at cruise Mach numbers.

Flow Around the Tips of a Three-Dimensional Wing

• Higher pressure on the lower surface of the wing relative tothe upper surface of the wing producing lift causes a flow ofair from the lower surface toward the upper surface around



A10, Page 15

air from the lower surface toward the upper surface, aroundthe wing tips.

• This flow has two effects:

– It causes a trailing vortex sheet that rolls up towards itsouter edges to form concentrated vortex cores.

– It causes an increased downward inclination to the airleaving the wing compared to an infinite span winggenerating the same lift.

Low Pressure

Flow Around the Tips of a Three-Dimensional Wing



A10, Page 16

+ ++ + + + + + +

+ + +

High Pressure

Leading Edge

Trailing Edge

General Flow Pattern Behind an AirplaneProduced by the Lift on the Wing



A10, Page 17

Vortex Sheet Shed by a 3-Dimensional Wing



A10, Page 18

Example of Vortex Sheet Impacting Clouds



A10, Page 19

• The production of lift by a three-dimensional wing resultsin one of the components of drag called induced drag.Induced drag is a result of the trailing vortex sheet

Induced Drag



A10, Page 20

Induced drag is a result of the trailing vortex sheet(downwash) produced by the wing.

• Lift is caused by a change in momentum of the airproduced by a turning of the airflow (Force = time rate ofchange of momentum).

• It can be shown that this momentum change is:

• The resultant reaction (for every action there is an equaland opposite reaction) is inclined aft relative to theoncoming airflow, producing Lift, along with anothercomponent we call induced drag.

22

V4bF ερπ=

Induced Drag

• It can be shown that the coefficient of induced dragcan be represented by the following equation:



A10, Page 21

• Induced drag varies directly with lift squared, andinversely with aspect ratio. (That is why high-performance sailplanes have wings with very highaspect ratios.)

AR = Aspect Ratio‘e’ is an efficiency factor e* AR*

CC2L

Dinduced π=

V C O S

Λ

V

V S I N Λ

ΛΛ

V

Effect of Wing Sweep



A10, Page 22

• Sweeping the wing has the effect of reducing the magnitude

of the component of the flow normal to the quarter-chord ofthe wing.

• Alternatively, it can be thought of as making the wing

appear thinner to the airflow passing over it.• Either way, the effect of sweeping the wing is to reduce the

amount by which the airflow velocities are accelerated asthe air flows around the wing.

C

no sweep

Effect of Wing Sweep on High-SpeedPerformance and Critical Mach Number

with sweep



A10, Page 23

CD

Mach

MCR

+.0020constant CL

• Sweeping the wing increases the critical Mach number,thus improving high-speed (cruise) performance, by

delaying the formation of shock waves to higher velocities.• The sudden increase in the wing’s drag is thus postponed,

allowing the airplane to operate efficiently at fastercruising speeds.

MDD MCR

MDD

Effect of Wing Sweep on High-SpeedPerformance and Critical Mach Number

0.08

0.100o Sweep

ea s i n g

S w e e

p



A10, Page 24

• In addition to delaying the critical Mach number, sweepingthe wing also delays the peak of the drag rise, and lowers

the magnitude of this peak.

CD

Mach

0.7 0.8 0.9 1.0

0

0.02

0.04

0.06

50o Sweep

35o Sweep

10o

Sweep I n

c r e



Video: “Approaching the Speed of Sound” (Continued)

(Part I of the Shell Oil Company series

“High Speed Flight”)

High-Speed Flow on a Wing

• At lower values of airspeed, the local velocity of flow overthe entire airfoil will be subsonic.

At higher airspeeds there will be supersonic flow over the



A10, Page 26

• At higher airspeeds, there will be supersonic flow over the

wing and the resulting formation of shock waves.• The lowest Mach number at which sonic velocity (M=1) is

attained on any part of an airfoil at the cruise lift coefficient isreferred to as the “Critical Mach number”. Sonic velocity atany point on an airfoil means that supersonic velocities willbe reached as the stream tube expands aft of the point ofsonic velocity.

• The flow must decelerate to subsonic speeds prior toreaching the trailing edge of the airfoil. This transition backto subsonic speeds usually occurs by the formation of ashock wave.


M=.50

Maximum Local Velocity

Equal to Sonic

Maximum Local Velocity

is Less than Sonic



A10, Page 27

M=.77

M=.72(Critical Mach Number)

M=.82

Equal to Sonic

SuperSonic

Flow

Normal Shock Wave

SubSonic Possible Separation

Separation

SuperSonicFlow

Normal Shock

Normal Shock


• A shock wave is a sharp pressure gradient, formed by allthe pressure pulses from downstream of the shock wave



A10, Page 28

the pressure pulses from downstream of the shock wave

piling up at the point where the velocity of flow is justdecelerated to Mach 1. Since pressure pulses can notmove upstream in a supersonic flow, the pulsesaccumulate at the point at which the flow is just sonic.

• Formation of shock waves on a wing results in anincrease of the wing’s drag. That drag increase rises veryrapidly after reaching the critical Mach number.

SHOCK WAVE

NEGATIVE

PRESSURE

MORE POSITIVE

PRESSURE AIRFLOW

DIRECTION

Effect of Shock Wave Formationon the Boundary Layer



A10, Page 29

B O U N D AR Y L AY E R

SU R F AC E O F AI R F O I L

Shock wave formation causes an increase in drag and a loss oflift because of:

– The energy loss through the shock wave.

– The adverse pressure gradient that exists across the shock.

– The increase in thickness of the boundary layer.

– The resulting turbulence region behind the shock.

Effect of Shock Wave Formationon the Pressure Distribution

Shock Wave



A10, Page 30

Fraction of Cord

UpperSurface

Lower

Surface

Low Mach Number

Cp

0 .2 .4 .6 .8 1.0

0

-

+

Fraction of Cord

UpperSurface

LowerSurface

Cp

0 .2 .4 .6 .8 1.0

0

-

+High Mach Number

Critical Pressure Coefficient

Effect of Mach Number on the Drag Polar

• At airspeeds below the critical Mach number, CD is essentially

constant at constant CL. At higher airspeeds, CD increases rapidly.

• The rapid increase in drag due to the formation of shock waves athigher Mach numbers is referred to as compressibility drag, CDM.



A10, Page 31

. 9

. 8

. 7

M = 0 t o

. 6

CL

CD

∆CDM

Mach

.2 .3 .4 .5 .6 .7 .8 .9

CROSS-PLOT OF CDM VS

MACH FOR A CONSTANT CL

g p y g, DM

Drag Analysis

• We have already introduced two components of the total

airplane drag. – Induced drag: Due to downwash and airflow around the

wingtips (drag due to lift).



A10, Page 32

g p ( g )

– Compressibility drag: Due to shock wave formation on thewing at higher speeds.

• There are several other components of drag; some can beinfluenced by airline operation, others are dictated by design.

– Skin friction drag

– Excrescence drag

– Interference drag

– Pressure drag

– Trim drag

– Other drag

Drag Analysis

• Skin Friction drag: The component of drag caused by the viscous

forces generated when air passes over the external surfaces ofthe airplane (i.e., friction!). This drag may be reduced byminimizing exposed surface areas and by promoting laminar flow

th i l



A10, Page 33

over the airplane.

• Excrescence drag: The component of drag caused by the sum ofall deviations from a smooth, sealed, external surface. Its growthcan be minimized by good maintenance practices:

– Mismatches and gaps (e.g., external patches, steps and gaps inskin joints and around windows, doors, access panels)

– Discrete items (e.g., antennas, masts, lights)

– Internal airflow and leakage (e.g., leaks of air from higherpressure to lower pressure surfaces due to deteriorated seals)

– Surface roughness (e.g., non-flush fasteners, rough paint andsurface finish, dirty airplane)

– Control surface mis-rigging

Drag Analysis

• Interference drag: The additional drag caused by the change in

flow pattern that accompanies the placing of two bodies inclose proximity; the drag of each item separately is less thanthe drag of the same items combined. e.g., the wing-bodyintersection can cause interference drag which can be reduced



A10, Page 34

intersection can cause interference drag which can be reduced

by using fillets. (The integration of the engine nacelles with thewing is a particularly difficult challenge for minimizinginterference drag).

• Pressure drag (a.k.a. Profile or Form): The component of dragcaused by the pressure distribution over the 3-dimensionalshape of the airplane. It can be reduced by careful shaping ofcritical areas such as the cockpit and aft body closure.

Includes pressure-induced airflow separation, but does notinclude Mach-induced separation.

Drag Analysis

• Parasite drag: The term used to denote all drag which is notrelated to either lift-induced or compressibility-induced drag:

Parasite drag =Skin friction drag +



A10, Page 35

Parasite drag =Skin friction drag +

Excrescence drag +Interference drag +

Pressure drag

• Trim drag: The component of drag due to deflection of thehorizontal stabilizer and elevator to trim the airplane for level,unaccelerated flight, plus the additional induced drag due tothe wing having to provide the additional lift required to offsetthe downward lift of the tail. This component of drag is oftenbook-kept as part of induced drag since it is really additionaldrag due to lift.

Drag Analysis

Other drag: Examples of additional sources of drag that can occur

at various times during flight include: – Landing gear extended drag

– Speedbrake (Spoiler) extension drag



A10, Page 36

Speedbrake (Spoiler) extension drag

– Engine inoperative drag: When calculating performance with anengine inoperative we have to include additional drag referred toas “windmill and spillage” drag.

– Deflected flight control surfaces drag: For example, when flying

with one engine inoperative, the asymmetry of thrust creates ayawing moment which must be compensated for by defection ofthe rudder to maintain straight, un-yawed flight. This rudderdeflection adds drag, as does the additional aileron deflection

required to offset the rolling moment caused by the rudderdeflection. This combined increase in drag is referred to as“control drag,” or “yaw drag.”

Drag Analysis

Total drag: The total drag of the airplane can beconsidered as the sum of all the pieces wehave discussed:



A10, Page 37

have discussed:

Total Airplane Drag = Skin friction drag +Excrescence drag +

Interference drag +

Pressure drag +Trim drag +

Induced drag +

Compressibility drag +Other drag

Parasite drag

Often group together as“induced and trim drag”



A10, Page 38

End of Aerodynamics (10):Wings

FLIGHT

OPERATIONS

ENGINEERING



A11, Page 1

Aerodynamics (11): Additional Aerodynamic Devices

Aerodynamics (11): Additional Aerodynamic Devices

• “High-Lift” Devices



A11, Page 2

• “Local – Flow” Improving Devices

• Wingtip Devices

• Speed Brakes/Spoilers

High-lift Devices

• Commercial jet airplane wings are designed to provide goodcruise performance, incorporating wing sweep, relatively thinwings, and low camber. The drawback of this design is a



A11, Page 3

relatively high stalling speed.

• High-lift devices are designed to increase the lifting capabilityof a wing at lower speeds, and thereby to reduce thestalling speed.

• Reduction of stalling speed is necessary in order to producelower takeoff and landing speeds, and thus acceptable takeoffand landing distances.

• High-lift devices come in numerous varieties and may becharacterized as trailing-edge flaps and leading-edge devices.

CL I n

c r e

a s i n

g F

l a p

CL

Flap Extended

Flap Retracted

Effect of Trailing Edge Flaps on Lift and Drag



A11, Page 4

α CD

• The effect of trailing edge flaps is to increase: the effective camber ofthe wing, the wing area, the maximum lift coefficient, and the lift at agiven angle of attack.

• Flaps also increase drag, requiring more thrust to maintain a givenspeed at at given weight. This drag increase can help in the control ofspeed and glide slope during an approach and landing. The engines areoperated at a higher, more responsive power setting.

• Flaps produce a lower approach attitude, and thereby improved visibilityof the runway area.

Plain Flap

Types of Trailing Edge Flaps



A11, Page 5

Split Flap

Fowler Flap

Slotted Flap


Plain Flap



A11, Page 6

Plain flap and split flap:

– Mechanically simple

– Low weight

– Low maintenance

– Less lift increase than other designs(lower aerodynamic efficiency)

Split Flap


Slotted Flap



A11, Page 7

Slotted flap relative to plain or simple flap: – Improvement over the plain or simple flap

– Mechanically more complex

– High-energy air from below the wing flowsthrough the slot and over the upper surfaceof the flap, delaying flow separation andimproving the lifting capability

– Improved aerodynamic efficiency


Fowler Flap



A11, Page 8

Fowler flap relative to the slotted flap: – Mechanically more complex

– Increased weight

– Provides greater lift (betteraerodynamic efficiency)

– Increases both the camber and thearea of the wing

CLSlotted

Fowler

Slotted

Fowler

Effect of Trailing Edge Flaps on Lift and Drag



A11, Page 9

α

PlainSplit

No Flap

CL

CD

Plain

Split

No Flap

Trailing Edge Flap Developments

Double or triple-slotted flaps:

– Further developments of theslotted and Fowler flaps

– Capable of very high liftDouble Slotted (Vane/Main) Flap



A11, Page 10

Capable of very high lift

coefficients – High mechanical complexity

(maintenance)

– Increased weight

– High drag at maximumdeflections

Double Slotted (Main/Aft) Fowler Flap

Triple Slotted Fowler Flap

Trailing Edge Flap Developments

Mid Fl

Fore Flaps

737-300/-400/-500

Triple-Slotted Flaps



A11, Page 11

Exhaust Gates

Aft Flaps

Mid Flap

Mid Flap

Aft Flaps

Main Flaps

737-600/-700/-800/-900

Continuous Span

Double-Slotted Flaps

slot closed

slot openslot

slat

CL

i

takeoff

landing

Leading Edge High Lift Devices



A11, Page 12

p

α

cruise

(contaminated)

• The goal of leading edge devices is to allow the wing to obtainhigher angles of attack before separation occurs.

• Leading edge devices increase maximum lift by:

– Delaying flow separation – Slightly increasing wing area

– Increasing CLMAX

– Lowering stall speeds

Types of Leading Edge Devices

Slotted Leading Edge Leading Edge Slat



A11, Page 13

S otted ead g dge

Drooped Leading Edge

Hadley Page Slat

ead g dge S at

Fixed Camber Krueger Flap

Variable Camber Krueger Flap

Typical Leading Edge DevicesUsed on Boeing Airplanes



A11, Page 14

Krueger Flap

Leading Edge Slat(Sealed or slotted)

(Plain or variable camber)

Flow-Improving Devices

• A variety of devices have been used to improve the‘local’ air flow over various parts of the airplane.



A11, Page 15

• Examples of flow-improving devices are:

– Leading edge fences

– Vortillons

– Vortex generators

– Nacelle chines

Leading Edge Fences



A11, Page 16

Leading edge fenceshave been used tocontrol undesirablespan-wise airflow.

Vortilons

Vortilons strategicallyplaced below the leadingedge of the wing can beused to help control the



A11, Page 17

p

airflow over aft areas ofthe upper surface ofthe wing.

(e.g., keeping it attachedover the ailerons at highangles of attack therebyimproving roll control.

Vortex Generators

Vortex

Direction of Flow

V t

Top View of

Generator



A11, Page 18

• Vortex generators are small wings of very low aspect ratio whichare placed on the body, engine nacelles, empennage, wing, etc.,at an angle to the local airflow.

• Vortex generators produce strong downstream vortices which can: – Energize the boundary layer, thereby reducing its thickness and

delaying separation

– Reduce spanwise flow and improve the wing lift distribution

Boundary Layer

Vortex

Generator

Vortex Generators

• VG’s can improve airplane performance by reducing theoverall airplane drag and improving control authority (i.e.,



A11, Page 19

handling qualities).• Can be used to reduce stall speeds and airframe vibration.

• Are commonly used on swept wings to alleviate “pitch up”

characteristics inherent in those wings.

• Are used wherever airflow separation is a problem (or couldpotentially be a problem).

Various Types of Vortex Generators

Outb’d

16 in.

Forwardedgeinspar skin

VG Planform

Vortex Generators



A11, Page 20

Vortex Generators

¾”3”

10.7” Nominal

VG Planform

Examples of Vortex Generator Locations onBoeing Airplanes

AIRPLANE LOCATION

707 Wing upper surface and under horizontal tail



A11, Page 21

727 Vertical tail, center engine inlet,and wing leading edge

737-200 Wing upper surface and aft body

adjacent to horizontal tail

737-300/400/500 Wing upper surface and engine nacelles

757 Wing upper surface

767 Wing upper surface and engine nacelles

Nacelle Chines

Nacelle chinesare large vortexgenerators, usedto control airflowseparation off ofthe nacelle



A11, Page 22

the nacelle

Nacelle Vortex Without Nacelle Chine

Boundary layerseparation caused byburst nacelle vortex

RetardingForce



A11, Page 23

Surface PressureDistribution

RetardingForce

NacelleVortex

Nacelle Vortex With Nacelle Chine

Strong vortex fromnacelle chine



A11, Page 24

Nacelle vortexcoalesces withchine vortex

Wingtip Devices—Winglets

• Winglets are basically small wings attached to thetips of an airplane’s wing and oriented at an angleto the wing



A11, Page 25

to the wing.

• There are a number of winglet designs currentlybeing used on commercial airplanes.

• Current Boeing commercial models incorporating

winglets are: – 747-400

– BBJ

– 737-700/800/900 (Optional) – MD-11

Winglets



A11, Page 26

747-400

Winglets

BBJ



A11, Page 27

MD-11

Winglet Effects

• The function of winglets is to reduce the strength ofthe wingtip vortex, to redistribute the lift across the



A11, Page 28

wing, and to thereby decrease induced drag.• Induced drag accounts for approximately 40

percent of total cruise drag. Any reduction inthis contribution to drag can produce substantial

fuel savings.

• The benefits of winglets are not free. Theirinstallation adds weight (bending moment) and skin

surface area.

Effect of Winglets on Wingtip Vortex Flow

Outboard flow on

lower surface

(A)

BEFORE



A11, Page 29

Vortex sheetrolls up to formconcentratedsheetSpillover from

lower surface toupper surface

Inboard flow onupper surface

Vortex

reduced

Winglet retards

spillover

Reduced inboardflow on upper

surface

Reduced outboard flowon lower surface

(B)

AFTER

Blended wingletwith smaller vortex

and less drag

Conventional wingtipwith large vortexand higher drag

Well-Designed Winglets Redistribute the LiftDistribution, Thereby Reducing Induced Drag



A11, Page 30

Wingtip Devices – Raked Tips

Revised slat(Outboard edge)

Newlights



A11, Page 31

The 767-400 incorporates a span increase (relative to

the –200 and –300) in the form of a raked tip insteadof a winglet. The end goal is the same: Reduction ofthe wingtip vortex and a redistribution of lift acrossthe wing.

5

6

g e C r u i s e

Drag Reduction of Wingtip Devices



A11, Page 32

Percentage Increase in Horizontal and Vertical Span

0 2 4 6 8 10 12 14 16 180

1

2

3

4

P e r c e n t a g e o f D r a g R

e d u c t i o n a t A v e r a g

KC-135 winglet

MD-11 winglet

747-400 tip plus winglet

737-800 blended winglet

767-400 raked tip

Other Lift-enhancing Devices

• Leading edge gloves

• Yehudi's



A11, Page 33

• Strakes and strakelets

Drag Devices: Speed Brakes and Spoilers

• The function of speed brakes and spoilers is toincrease drag, and decrease lift by interfering with



A11, Page 34

the airflow over the wing.• Spoilers/speed brakes are used to increase rates of

descent, and to improve deceleration during landingor a rejected takeoff.

• During landing, or a rejected takeoff, the spoilersnot only add drag, but also place a greater amountof weight on the landing gear (due to the loss of lift),

enhancing braking effectiveness.

Spoiler / speed brakepanels (both sides)

Spoilers/Speed Brakes



A11, Page 35

Spoilersdeployed

f l a p

Effect of Spoilers / Speed Brakes on Lift



A11, Page 36

CL

α

I n c r e

a s i n

g

f

I n c r e

a s i n

g

s p o i l e

r s



A11, Page 37

End of Aerodynamics (11): Additional Aerodynamic Devices

FLIGHT

OPERATIONS

ENGINEERING



A12, Page 1

Aerodynamics (12):Flight Control Devices

Aerodynamics (12): Flight Control Devices

• Primary

• Trim



A12, Page 2

• High Lift

Flight Control Devices

AILERONSRUDDER

ELEVATOR



A12, Page 3

LEADING EDGE SLATS

TRAILING EDGE FLAPS

SPOILERS

HORIZONTAL

STABILIZER

Primary Flight Controls

• Pitch axis

– Elevators



A12, Page 4

• Roll axis

– Ailerons (high and low speed)

– spoilers

• Yaw axis

– Rudder

– Yaw damper

Trim Controls

• Pitch trim

– Stabilizer/elevator trim



A12, Page 5

• Roll trim

– Aileron trim

• Yaw trim

– Rudder trim

High-lift Devices

• Leading edge flaps, slats

• Trailing edge flaps



A12, Page 6



A12, Page 7

End of Aerodynamics (12):Flight Control Devices

FLIGHT

OPERATIONS

ENGINEERING



A13, Page 1

Aerodynamics (13):Data Presentation in the PEM

Aerodynamics (13):Data Presentation in the PEM

• Brief review of 7X7 PEM Sections 1 and 2

• Calculating the Aerodynamic Performance of an



A13, Page 2

Actual Airplane



A13, Page 3

Brief Review of 7X7 PEM

Sections 1 and 2

Aerodynamic Performance of an Actual Airplane

• In previous topics we have defined and illustrated, ingeneral terms, some of the basic aerodynamiccharacteristics of airplanes. Let us now look at the data



A13, Page 4

for an actual airplane. The airplane we will use as ademonstration is the Boeing 7X7.

• The answers to the questions that follow may be

derived from the charts found in your copy of the 7X7Performance Engineer’s Manual.

• Note, from page 1.6 of the PEM, that the reference

wing area, S, for the 7X7 is 1341 square feet.

Question #1: (Lift and Drag)



A13, Page 5

A 7X7 is in steady-state climb after takeoff withthe flaps and gear up. The airplane weighs160,000 lb, is at a pressure altitude = 5000 feet,and the equivalent airspeed is 250 knots.

Assuming lift = weight, what is the value of CL?




A13, Page 6

For the same conditions as Question #1, what isthe value of CD, and what is the total drag force?

(Refer to PEM page 2.5)




A13, Page 7

What is the body angle of attack (alpha) for thevalue of CL determined in Question #1?





A13, Page 8

A 7X7 is flying at Mach .79, at an altitude of33,000 ft. The weight of the airplane is 140,000pounds. What is the airplane’s CL and the angle

of the cabin floor relative to horizontal?(assume lift = weight)



The climb gradient of an airplane may be approximated as:

(You will be learning a more precise form of this equationl t i th )

−

CL

CD

weightthrust

Climb gradient, γ =



A13, Page 9

later in the course.)

Rate of climb, in feet per minute, can be approximated by:

Suppose a 7X7 is climbing after takeoff, flaps and gear areup, both engines are operating. Weight = 160,000 pounds,pressure altitude = 10,000 feet, OAT = ISA, Mach = .450.Thrust per engine = 12,800 pounds. Assume lift = weight,and neglect the Reynold’s number correction to drag.

(Continued on next page)

ROC = γ * VKTAS * 101.268




A13, Page 10

5a) Calculate the climb gradient and the rate of climb.





A13, Page 11

5b) Now, assume the speedbrakes are fullyextended. Calculate the new climb gradientand rate of climb.


Question #6: (Reynolds Number Correction)



A13, Page 12

A 7X7 is in steady-state cruise at FL370. The weightis 58,967 kilograms, the cruise Mach = .79, and thetemperature is ISA + 20° C.

6a) Calculate the Reynolds number correction to thedrag coefficient using the chart on PEM page 2.16.

Question #6: (Reynolds Number Correction)



A13, Page 13

6b) Now, calculate the drag force with and withoutthe Reynolds number correction included.

(Refer to PEM pages 2.13 and 2.16)

Question #7: (Initial Buffet)



A13, Page 14

Find the stall speed (in knots calibratedairspeed) for a 7X7 weighing 140,000 pounds;10,000 ft; standard day; flaps up.


Compare that result to PEM page 2.28

E d f A d i (13)



End of Aerodynamics (13):Data Presentation in the PEM

Date post:	06-Jul-2018
Category:	Documents
Upload:	bnolasco
View:	215 times
Download:	0 times

Basic Aircraft.engineers.pvsbico.aerodynamicsd

Documents