ae172lecturenoteskutay

Middle East Technical University Department of Aerospace Engineering AE172 Lecture Notes Spring 2010 Dr. Ali Türker Kutay

1

Forces and Moments Acting on an A/C

Airfoil The shape of an aerodynamic force generating surface (such as wing, propeller blade, A/C control surface,

sail, etc.) as seen in cross-section obtained by intersecting the wing with a perpendicular plane.

Chord line: a straight line connecting the leading and trailing edges of an airfoil.

Mean camber line: a line drawn midway between the upper and lower surfaces.

Note: For a symmetric airfoil the chord and mean camber lines coincide.

Chord: length of the chord line, denoted by 𝑐.

Camber: the distance between the mean camber line and chord line, usually represented as a percentage

of the chord.

Upper surface (suction side)

Lower surface (pressure side)𝑉∞ 𝛼


2

Thickness: the distance between the upper and lower surfaces, usually represented as a percentage of the

chord.

𝑽∞ : speed of the oncoming flow (if there is no wind, it is the same as the speed of the A/C, but in the

opposite direction).

Angle of attack: the angle between the chord line and the oncoming flow, represented by 𝛼. Positive if the

flow is below the chord.

NACA Airfoils Airfoils developed by National Advisory Committee for Aeronautics (NACA) of USA. In 1958 NACA was

replaced by NASA (National Aeronautics and Space Administration).

The shape of the NACA airfoils is described using a series of digits. The parameters in the numerical code

can be entered into equations to precisely generate the cross-section of the airfoil.

Four-digit series

NACA xxxx

1. First digit describes the maximum camber as percentage of the chord.

2. Second digit describes the distance of maximum camber from the airfoil leading edge in tens of

percents of the chord.

3. The last two digits describe the maximum thickness of the airfoil as percent of the chord.

Example:

The NACA 2412 airfoil has a maximum camber of 0.02𝑐 located at 0.4𝑐 from the leading edge with a

maximum thickness of 0.12𝑐.

The NACA 0015 airfoil is symmetrical, the 00 indicating that it has no camber. The 15 indicates that the

airfoil has a maximum thickness of 0.15𝑐.

Aerodynamic Forces Aerodynamic forces are created by the pressure distributions over the bodies. Every part of an A/C

exposed to air is under pressure and hence contributes to aerodynamic forces. This is why the landing

gears are extracted during flight, and nacelles and pylons are used to make various components

aerodynamically efficient.

Forces on an airfoil


3

The net effect of the distributed pressure can be exactly represented by a single concentrated force 𝑅. The

resultant force 𝑅 (its magnitude, direction, and the point where it applies on the chord, i.e., center of

pressure (c.p.)) depends on various factors, including 𝑉∞ , 𝛼, 𝑐, and camber and thickness distribution. The

component of 𝑅 in the direction of the flow is called drag (𝐷) and the component perpendicular to flow is

called lift (𝐿).

If we place an axis perpendicular to the airfoil that passes through the c.p. there will be no moment about

the axis. Therefore c.p. may also be defined as the point on the chord about which the moment is zero.

From the principles of statics 𝑅 can be shifted to any arbitrary point, together with the moment about that

point.

Similarly there is a single resultant force vector 𝑅 for the wing, and another the entire A/C. The c.p. for the

A/C is different than the c.p. of the airfoil.

Symmetric airfoils (no camber)

For 𝛼 = 0 the flow is symmetric and hence 𝐿 = 0.

Stramline: The path of a fluid particle in steady flow. Visible particles (smoke) are injected into flow to

make streamlines visible in wind tunnel tests.

For lift to be created nonzero 𝛼 is required.


4

For 𝛼 > 0, 𝐿 > 0 and for 𝛼 < 0, 𝐿 < 0.

Flow turning at trailing edge is very important. Fluid particles change direction (accelerate downwards) as

they pass along the airfoil. The wing applies a force on air particles to accelerate them downwards. By the

action-reaction principle (Newton’s 3rd law of motion) flow particles apply the same force back on the wing

in the opposite direction. This creates lift and drag.

This is how control surfaces work. Flow turning (and hence lift force) is controlled by controlling the

deflection angle of the control surfaces at the trailing edge.

Note: For thin symmetric airfoils center of pressure lies close to 25% of the chord behind the leading edge.

This is called the quarter-chord point. As 𝛼 is increased, 𝑅 grows in magnitude, but center of pressures

does not move.

Cambered airfoils

With the help of camber, flow turning at trailing edge (and hence positive𝐿) is obtained even at 𝛼 = 0.

Unlike symmetric airfoils, c.p. for cambered airfoils move with 𝛼.


5

Aerodynamic Center

The point on the airfoil about which moments do not vary with 𝛼.

It has been found both experimentally and theoretically that on most low speed airfoils the a.c. is very close

to the c/4 point.

For symmetric airfoils both the c.p. and a.c. are fixed at c/4.

For cambered airfoils c.p. moves with 𝛼, but a.c. is fixed at c/4.

𝑅3 < 𝑅2 < 𝑅1

Aerodynamic Coefficients 𝑅 (𝐿 and 𝐷) Changes with Relation

Density (𝜌∞ , kg/m3) Linear

Wing size (planform area) (𝑆, m2) Linear

Air speed (𝑉∞ , m/s) Quadratic

Angle of attack (𝛼, rad) Nonlinear (linear for low 𝛼)

Airfoil shape (camber, thickness) Nonlinear

Wing shape (aspect ratio, taper, etc.) Nonlinear

Viscosity and compressibility effects Nonlinear

Based on the above, aerodynamic forces and moments are written as

𝐿 =1

2𝜌∞𝑉∞

2

𝑞∞

𝑆𝐶𝐿

𝐷 =1

2𝜌∞𝑉∞

2

𝑞∞

𝑆𝐶𝐷

𝑀 =1

2𝜌∞𝑉∞

2

𝑞∞

𝑆𝑐 𝐶𝑀


6

The effects of all the complex nonlinear dependencies are collected in nondimensional coefficients

𝐶𝐿 = 𝐶𝐿 𝛼, 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑐𝑎𝑚𝑏𝑒𝑟,𝑀∞ ,𝑅𝑒 , 𝐶𝐷, 𝐶𝑀, ….

𝐶𝐿 =𝐿

𝑞∞𝑆

𝐶𝐷 =𝐷

𝑞∞𝑆

𝐶𝑀 =𝑀

𝑞∞𝑆𝑐

The aerodynamic coefficients are usually determined experimentally and then used to predict forces and

moments.

Viscosity: describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction.

Accounted for using a dimensionless number called the Reynolds number (Re).

Reynolds number (Re): is a dimensionless number that gives a measure of the ratio of inertial forces to

viscous forces and consequently quantifies the relative importance of these two types of forces for given

flow conditions.

Re =𝜌∞𝑉∞𝑐

𝜇

𝜇: coefficient of dynamic viscosity, a property of fluid.

Typical Reynolds Numbers:

Full scale airliner above 10 000 000

Light aircraft above 1 000 000

Large model aircraft less than 400 000

Typical model aircraft less than 200 000

Indoors and slow flyers less than 30 000

Compressibility: is a measure of the relative volume change of a fluid as a response to a pressure change.

Becomes a problem at high speeds. Accounted for using a dimensionless number called the Mach number

(M).

M =𝑉∞𝑎∞

𝑎∞ : speed of sound.


7

Lift coefficient

Stall: is a condition in aerodynamics where the angle of attack increases beyond a certain point such that

the lift begins to decrease.

The 𝐿 and 𝐷 decomposition of 𝑅 was demonstrated on a 2D airfoil. In 2D analysis we assume that we have

a wing with infinite span and the forces are the same at each spanwise section.


8

Small case letters are used for 2D analysis (𝐶𝑙 ,𝐶𝑑 ,𝐶𝑚 ,…).

Things get complicated for 3D aircraft because the forces are not the same (𝛼 is not the same) everywhere

along the span.

For given aerodynamic characteristics of an airfoil (𝐶𝑙 ,𝐶𝑑 ,𝐶𝑚 ,…), the aerodynamic performance of a wing

depends on the aspect ratio and planform.

𝐴𝑅 =𝑏2

𝑆

Drag coefficient

For finite wings the flow spillage from lower surface to top at wing tips lowers lift and increases drag. The

increase in drag is called the “induced drag” and is a function of lift. When there is no lift there is no flow

spillage and no additional drag.

𝐶𝐷 = 𝐶𝐷0 parasitic drag

(form drag ,or

zero −lift drag )coefficient

+𝐶𝐿

2

𝜋𝑒𝐴𝑅 induced drag

coefficient

(for subsonic flows)

𝑒: Oswald (span) efficiency number, a constant that represents the lifting efficiency of a three dimensional

wing compared with another wing having the same AR and an elliptical lift distribution.


9

𝑒 =1

1 + 𝛿

http://en.wikipedia.org/wiki/File:Drag_Curve_2.jpg


10

Drag polar

𝐶𝐷 = 𝐶𝐷0 +𝐶𝐿

2

𝜋𝑒𝐴𝑅

Standard Atmosphere Air temperature and pressure depends on altitude, location, time of day, season, solar activities, etc. It is

impossible to take into account all these factors when designing and analyzing flight vehicles. Standard

atmospheres are defined to form a common reference for mean values of temperature, density, pressure,

etc. as functions of altitude.

Altitude

𝑕𝐺 : Geometric altitude, geometric height above sea level

𝑕𝑎 : Absolute altitude, from the center of the Earth. Important for space flight since gravity changes with

𝑕𝑎 .

𝑕𝑎 = 𝑕𝐺 + 𝑟

𝑟: radius of the Earth

𝑔 = 𝑔0 𝑟

𝑕𝑎

2

= 𝑔0 𝑟

𝑕𝐺 + 𝑟

2

𝑔0: gravitational acceleration at sea level


11

In altitudes where most A/C fly (𝑕𝐺 < 20 km) change in 𝑔 is small. As an approximation we write (1) as

𝑑𝑝 = −𝜌𝑔0𝑑𝑕 (2)

Equations (1) and (2) give close results. They can be made exactly the same by introducing a fictitious

altitude that is slightly different than 𝑕𝐺 , to make future calculations simpler.

𝑕: Geopotential altitude, a fictitious altitude physically compatible with the assumption that

𝑔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑔0.

−𝜌𝑔 𝑕𝐺 𝑑𝑕𝐺 = −𝜌𝑔0𝑑𝑕

𝑑𝑕 =𝑔

𝑔0𝑑𝑕𝐺

=𝑟2

𝑟 + 𝑕𝐺 2𝑑𝑕𝐺

Integrating gives

𝑕 = 𝑟

𝑟 + 𝑕𝐺 𝑕𝐺

Atmosphere Layers

Experimental evidence shows that the temperature variation with altitude consists of series of straight

lines.

𝑎 =𝑑𝑇

𝑑𝑕

𝑑𝑕𝐺

1

1

𝑝 + 𝑑𝑝

𝑝 𝑕𝐺

𝑕𝐺

𝑊

𝑊 = 𝜌 1 1 𝑑𝑕𝐺 𝑣𝑜𝑙𝑢𝑚𝑒 𝑚𝑎𝑠 𝑠

𝑔 𝑕𝐺

𝑊: weight of the fluid element

Force balance on the fluid element:

𝑝 = 𝑝 + 𝑑𝑝 + 𝜌𝑔𝑑𝑕𝐺

𝑑𝑝 = −𝜌𝑔 𝑕𝐺 𝑑𝑕𝐺 (1)


12

Ideal gas law:

𝑝 = 𝜌𝑅𝑇 (3)

𝑝: pressure

𝑇: absolute temperature

𝜌: density

𝑅: = 287J

Kg K , specific gas constant

Based on the approximated temperature variation 𝑝 and 𝜌 can be found using equations (2) and (3).

Results are presented as standard atmosphere tables.

Pressure, Temperature, and Density Altitudes

Altitudes in the standard atmosphere table corresponding to measured pressure, temperature, and density

values.

Example: An aircraft flying at a certain altitude measures the temperature and pressure as

𝑝 = 4.72 × 104𝑁/𝑚2

𝑇 = −17.45 °𝐶 = 255.7 𝐾

The density can be found from

𝜌 =𝑝

𝑅𝑇=

4.72× 104

287 × 255.7= 0.6432 𝑘𝑔/𝑚3

According to 1959 US standard atmosphere, these values correspond to:

-100 -50 0 50 100 1500

11

20

32

4751

71

84.852

Temperature (C)

Geopote

ntial A

ltitude (

km

)

a1 = -6.5 oC/km Troposphere

a2 = 0 oC/km Tropopause

a3 = +1.0 oC/km Stratosphere

a4 = +2.8 oC/km

a5 = 0 oC/km Stratopause

a6 = -2.8 oC/km Mesosphere

a7 = -2.0 oC/km


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𝑕𝑝 = 6,000 𝑚

𝑕𝑇 = 5,000 𝑚

𝑕𝜌 = 6,240 𝑚

The actual altitude of the aircraft is yet a different number!

Forces and Moments on an A/C in Longitudinal (Vertical, Pitching) Plane No lateral motion (no rolling and yawing)!

Forces parallel to flight path:

𝐹∥ = 𝑇 𝑐𝑜𝑠 𝛼 − 𝐷 −𝑊𝑠𝑖𝑛 𝛾 = 𝑚𝑎 = 𝑚𝑑𝑉

𝑑𝑡

Forces perpendicular to flight path:

𝐹⊥ = 𝐿 + 𝑇 sin𝛼 −𝑊 cos 𝛾 = 𝑚𝑉∞

2

𝑟𝑐: centrifugal acceleration

Special Cases

Horizontal flight without acceleration

𝛾 = 0,𝑑𝑉

𝑑𝑡= 0

Then

𝑇 = 𝐷

𝐿 = 𝑊

Aircraft Performance What can an aircraft do?

- How fast can it fly?

- What is the maximum altitude it can fly at? (ceiling)

- How far can it go? (range)

- How long can it stay in the air? (endurance)

- How fast can it climb to the cruising altitude?

- How fast can it descend?

- How quickly can it stop on the ground?

- How quickly can it take off?

𝑉∞

𝛾 𝛼 𝜃

𝐷

𝐿

𝑇

𝑊

𝜃: pitch angle 𝛾: flight path angle 𝛼: angle of attack


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- What is the minimum radius of turn?

- How many g’s can an aircraft pull?

Thrust Required for Level Unaccelerated Flight 𝑇𝑅: Thrust required to maintain the given flight condition (speed and altitude). Depends on the velocity,

altitude, shape, size, and weight of the A/C. Has nothing to do with the engines.

𝑇𝑅 = 𝐷 = 𝑞∞𝑆 𝐶𝐷

=𝑊

𝐶𝐿

𝐶𝐷

∴ 𝑇𝑅 =𝑊

𝐶𝐿/𝐶𝐷=

𝑊

𝐿/𝐷

Thrust required, 𝑇𝑅, for a given airplane at a given altitude, varies with velocity, 𝑉∞ . This variation may be

plotted as follows:

1. Choose a value of 𝑉∞

2. For this 𝑉∞ calculate 𝐶𝐿 from

𝐶𝐿 =𝑊

12𝜌∞𝑉∞

2𝑆

3. Calculate 𝐶𝐷 from the drag polar

𝐶𝐷 = 𝐶𝐷0 + 𝐾𝐶𝐿2

4. Calculate the ratio 𝐶𝐿/𝐶𝐷

5. Then

𝑇𝑅 =𝑊

𝐶𝐿/𝐶𝐷

𝑇𝑅 ∝1

𝐶𝐿/𝐶𝐷

Thrust varies inversely with 𝐿/𝐷. Therefore minimum 𝑇𝑅 will be obtained at 𝑉∞ where 𝐿/𝐷 is maximum. A

typical 𝑇𝑅 curve looks like


15

The lift-to-drag ratio, 𝐿/𝐷, is a measure of aerodynamic efficiency. It is a function of 𝛼, it reaches a

maximum at some specific 𝛼.

Different points on the 𝑇𝑅 curve correspond to different angles of attack.

𝑇 = 𝐷 = 𝑞∞𝑆 𝐶𝐷0 + 𝐾𝐶𝐿2

= 𝑞∞𝑆𝐶𝐷0 𝑧𝑒𝑟𝑜 −𝑙𝑖𝑓𝑡 𝑇𝑅

+ 𝑞∞𝑆𝐾𝐶𝐿2

𝑇𝑅 𝑑𝑢𝑒 𝑡𝑜 𝑙𝑖𝑓𝑡

Note: At minimum thrust required

𝑇𝑅

𝑉∞

Zero-lift 𝑇𝑅

𝑇𝑅 due to lift

𝑉 𝐿/𝐷 𝑚𝑎𝑥

𝑇𝑅

𝑉∞

Increasing velocity

Decreasing 𝛼

𝐿/𝐷

𝛼

𝛼 𝐿/𝐷 𝑚𝑎𝑥

𝑇𝑅 𝑁

𝑉∞ 𝑚/𝑠 𝐿/𝐷 𝑚𝑎𝑥

better

typical


16

𝐶𝐷0 = 𝐶𝐷𝑖

zero lift drag = drag due to lift

𝐿 = 𝑊 =1

2𝜌∞𝑉∞

2𝑆𝐶𝐿 ⇒ 𝐶𝐿 =2𝑊

𝜌∞𝑉∞2𝑆

(Lift eqn)

𝑇𝑅 = 𝐷 =1

2𝜌∞𝑉∞

2𝑆 𝐶𝐷0+ 𝐾𝐶𝐿

2

=1

2𝜌∞𝑉∞

2𝑆 𝐶𝐷0+ 4𝐾

𝑊

𝜌∞𝑉∞2𝑆

2

=1

2𝜌∞𝑉∞

2𝑆𝐶𝐷0+

2𝐾𝑆

𝜌∞𝑉∞2 𝑊

𝑆

2

= 𝑞∞𝑆𝐶𝐷0+𝐾𝑆

𝑞∞ 𝑊

𝑆

2

We can solve for 𝑉∞ from the above equation as

𝑉∞2 =

𝑇𝑅𝑆 ±

𝑇𝑅𝑆

2

− 4𝐶𝐷0𝐾

𝑊𝑠

2

𝜌∞𝐶𝐷0

Not a new equation, obtained from (Lift) and (Drag) equations!

𝑇𝑅𝑆

=𝑇𝑅𝑊

𝑊

𝑆

𝑉∞ =

𝑇𝑅𝑊𝑊𝑆 ±

𝑊𝑆 𝑇𝑅𝑊

2

− 4𝐶𝐷0𝐾

𝜌∞𝐶𝐷0

1 2

(𝑉∞ eqn)

For a given 𝑇𝑅, speed of an A/C depends on:

1. Thrust-to-weight ratio, 𝑇𝑅/𝑊

2. Wing loading, 𝑊/𝑆

3. The drag polar, 𝐶𝐷0 and 𝐾

These are the fundamental parameters that dictate A/C performance.

Circular argument in aircraft design:

𝑇

𝑊: Thrust-to-weight ratio, a dimensionless indicator of engine performance.

𝑊

𝑆: In cruise airplanes fly at much higher velocities than takeoff and landing and the necessary lift can be

created with smaller wings. Extra wing area required for takeoff and landing is extra load for cruise.

Carry more weight

Fly faster

Bigger engine (heavier)

Solution: Increase thrust without adding weight


17

Thrust Available 𝑇𝐴: thrust provided by the engine(s). Strictly associated with the engine. Can vary with 𝑉∞ , 𝜌∞ depending

on the type of the engine.

Maximum Velocity Intersection of the 𝑇𝑅 curve and the maximum 𝑇𝐴

curve defines the maximum velocity 𝑉𝑚𝑎𝑥 of the

aircraft at the given altitude. This is an important

aspect of the aircraft design process.

𝑇𝑅 𝑉𝑚𝑎𝑥 = 𝑇𝐴 𝑚𝑎𝑥

𝑉𝑚𝑎𝑥 can be found by inserting 𝑇𝐴 𝑚𝑎𝑥 into (𝑉∞ eqn)

Power Required 𝑃𝑜𝑤𝑒𝑟 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒

=𝑓𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 𝑓𝑜𝑟𝑐𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑃𝑅 = 𝑇𝑅𝑉∞ , 𝑁.𝑚

𝑠 = 𝑊 = 𝑊𝑎𝑡𝑡

=𝑊

𝐶𝐿𝐶𝐷

𝑊

12𝜌∞𝑆𝐶𝐿

=𝑊

𝐶𝐿3 2

𝐶𝐷

2𝑊

𝜌∞𝑆

Then

𝑃𝑅 ∝1

𝐶𝐿3 2 /𝐶𝐷

Note: 746 𝑊 = 1 𝑕𝑝 (horse power)

Note: The slope of the line from origin to any point on

the 𝑃𝑅 vs 𝑉∞ curve is 𝑇𝑅

𝑃𝑅𝑉∞

=𝑇𝑅𝑉∞𝑉∞

= 𝑇𝑅

40 60 80 100 1201500

2000

2500

3000

3500

V (m/s)

TR

(N

)

TA

(partial)

TA

(max)

V(max)

0 20 40 60 80 100 1200

0.5

1

1.5

2

2.5

3

3.5x 10

5

PR

(W

)

V(min P

R)

V(min T

R)

(CL

3/2/C

D)max

(CL/C

D)max


18

Power required:

𝑃𝑅 =1

2𝜌∞𝑉∞

2𝑆 𝐶𝐷0+ 𝐾𝐶𝐿

2 𝑉∞, 𝐶𝐿 =𝑊

12𝜌∞𝑉∞

2𝑆

𝑃𝑅 =1

2𝜌∞𝑉∞

3𝑆𝐶𝐷0+

2𝐾𝑊2

𝜌∞𝑉∞𝑆

Minimum 𝑃𝑅:

𝑑𝑃𝑅𝑑𝑉∞

=3

2𝜌∞𝑉∞

2𝑆𝐶𝐷0−

2𝐾𝑊2

𝜌∞𝑉∞2𝑆

= 0

3𝐶𝐷0= 𝐾

𝑊

12𝜌∞𝑉∞

2𝑆

2

= 𝐾𝐶𝐿2

The aerodynamic condition that holds at 𝑃𝑅 𝑚𝑖𝑛 : ∴ 𝐶𝐷𝑖 = 3𝐶𝐷0

Power Available 𝑃𝐴: power provided by the engine(s). Strictly associated with the engine. Can vary with 𝑉∞ , 𝜌∞

depending on the type of the engine.

Propulsive Characteristics A propulsion system is a machine that produces thrust to push an A/C forward. Thrust is usually generated

through application of Newton's third law of action and reaction. Air is accelerated by the engine, and the

reaction to this acceleration produces a force on the engine. The amount of thrust generated depends on

the mass flow through the engine and the exit velocity of the air.

Thrust produced is given by Newton’s 2nd law of motion (force applied is equal to the rate of change of

momentum) as

𝑇 =𝑑 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚

𝑑𝑡=𝑑 𝑚 𝑉𝑗 − 𝑉∞

𝑑𝑡

= 𝑚 𝑉𝑗 − 𝑉∞

𝑚 : mass flow rate through the propulsion device

Propulsion device

𝑉∞ ,𝑝∞ 𝑉𝑗 ,𝑝𝑗 𝑇

Force exerted by the air on the propulsive device

Equal and opposite force exerted by the propulsive device on the air


19

The exiting air accelerated from 𝑉∞ to 𝑉𝑗 has the following kinetic energy per unit mass:

1

2 𝑉𝑗 − 𝑉∞

2

This kinetic energy is totally wasted. The total power produced by the engine is:

𝑃𝑇 = 𝑇𝑉∞ useful to

push the A/Cforward

+1

2𝑚 𝑉𝑗 − 𝑉∞

2

wasted

Efficiency of the propulsion system is defined as:

𝜂𝑃 =useful power available

total power generated

=𝑇𝑉∞

𝑇𝑉∞ +12𝑚 𝑉𝑗 − 𝑉∞

2

Using 𝑇 = 𝑚 𝑉𝑗 − 𝑉∞ we get

𝜂𝑃 =2

1 +𝑉𝑗𝑉∞

Propulsion alternatives:

- Reciprocating engine – propeller combination

- Turbojet

- Turbofan

- Turboprop

Propeller: Large diameter, moves large mass of air, but gives air small acceleration. High efficiency,

low thrust. Thrust is limited by propeller tip speed.

Jet Engine: Small diameter, moves small amount of air, but gives air high acceleration. Low

efficiency, high thrust.

Reciprocating Engine - Propeller Combination

The motion of one or more pistons is transformed into the rotary motion of a crankshaft, which transmits

the engine’s power to air through a propeller.

𝑃 is reasonably constant with 𝑉∞ .

Propeller: is a twisted wing attached vertically to the longitudinal axis of the airplane and produces lift and

drag (skin friction, pressure, induced, wave (compressible)). Drag is a loss that extracts from the useful

power.

𝑃𝐴 (available power, transmitted to air) < 𝑃 (shaft power produced by the engine)

𝑃𝐴 = 𝜂𝑃𝑟𝑃

𝜂𝑃𝑟 : propeller efficiency ≈ 0.7− 0.8.

Turbojet Engine

For a turbojet engine flying at subsonic speeds 𝑇 can be considered constant with 𝑉∞ .


20

Output of a turbojet engine is measured in terms of thrust, while the output of a piston engine is measured

in terms of power.

Example: Consider the following A/C:

𝑏 = 10.91 𝑚

𝑆 = 16.17 𝑚2

𝑊 = 13139 𝑁

𝑃𝑚𝑎𝑥 = 230 𝑕𝑝 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙

= 230 𝑕𝑝 × 746𝑊

𝑕𝑝

= 171580 𝑊

𝐶𝐷0= 0.025

𝑒 = 0.8

𝜂𝑃𝑟 = 0.8

a) Calculate the thrust and power required at sea level and 𝑉∞ = 61 𝑚/𝑠.

b) Calculate the maximum speed at sea level.

Solution: a)

𝐿 = 𝑊 ⇒ 𝐶𝐿 =𝑊

12𝜌∞𝑉∞

2𝑆=

13139

12 × 1.225× 612 × 16.17

= 0.357

𝐶𝐷 = 𝐶𝐷0+

𝐶𝐿2

𝜋𝑒𝐴𝑅= 0.025 +

0.3572

3.14 × 0.8×10.912

16.17

= 0.0319

Then

𝐶𝐿𝐶𝐷

=0.357

0.0319= 11.19

𝑇𝑅 =𝑊

𝐶𝐿𝐶𝐷

=13139

11.19= 1174.5 𝑁

𝑃𝑅 = 𝑇𝑅𝑉∞ = 1174.5× 61 = 71647 𝑊 = 96 𝑕𝑝

b) Maximum speed occurs when 𝑃𝑅 = 𝑃𝐴 𝑚𝑎𝑥

𝑃𝐴 𝑚𝑎𝑥 = 𝜂𝑃𝑃𝑚𝑎𝑥 = 0.8 × 171580 = 137264 𝑊


21

Altitude Effects on Power and Thrust Lower air density at higher altitudes causes a reduction in power for both the reciprocating and jet engines.

In this course we will assume 𝑃𝐴 and 𝑇𝐴 to be proportional to air density.

𝑇𝐴,𝑎𝑙𝑡 =𝜌𝑎𝑙𝑡𝜌0

𝑇𝐴,0

𝑃𝐴,𝑎𝑙𝑡 =𝜌𝑎𝑙𝑡𝜌0

𝑃𝐴,0

where the subscript “0” indicates sea level.

Example: Consider the A/C in the last example.

𝑏 = 10.91 𝑚

𝑆 = 16.17 𝑚2

𝑊 = 13139 𝑁

𝑃𝑚𝑎𝑥 = 230 𝑕𝑝 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙

= 230 𝑕𝑝 × 746𝑊

𝑕𝑝

= 171580 𝑊

𝐶𝐷0= 0.025

𝑒 = 0.8

𝜂𝑃𝑟 = 0.8

Suppose that 𝐶𝐿𝑚𝑎𝑥 = 1.5. Then

𝑉𝑠𝑡𝑎𝑙𝑙 = 2𝑊

𝜌0𝑆𝐶𝐿𝑚𝑎𝑥= 29.7

𝑚

𝑠 @ 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙

𝑃𝑅(𝑉∞) =

𝑊𝐶𝐿𝐶𝐷

𝑉∞ =𝑊

𝑊

12𝜌∞𝑉∞

2𝑆

𝐶𝐷0+ 4𝐾

𝑊𝜌∞𝑉∞

2𝑆

2

𝑉∞

𝑃𝐴 𝑚𝑎𝑥 = 𝜂𝑃𝑃𝑚𝑎𝑥 = 0.8 × 171580 = 137264 𝑊

𝑉𝑚𝑎𝑥 = 79.7𝑚

𝑠 @ 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙

Consider the altitude where 𝜌1 = 0.52 𝑘𝑔/𝑚3 (~𝑕𝐺 = 8095 𝑚).



𝑚

𝑠 @ 8095 𝑚

𝑃𝐴 𝑚𝑎𝑥 = 𝜂𝑃𝜌𝑎𝑙𝑡𝜌0

𝑃𝑚𝑎𝑥 = 0.8 ×0.52

1.225× 171580 = 58267 𝑊

𝑉𝑚𝑎𝑥 = 56.6𝑚

𝑠 @ 8095 𝑚

𝑉𝑚𝑖𝑛 = 46.6𝑚

𝑠> 𝑉𝑠𝑡𝑎𝑙𝑙 @ 8095 𝑚

0 20 29.7 45.6 56.6 79.70

5.8267

13.7264

x 104

V (m/s)

PR

(W

)

(PA)max

@ 8095 m

(PA)max

@ 0 m


22

At sea level 𝑉𝑚𝑖𝑛 = 𝑉𝑠𝑡𝑎𝑙𝑙

At 8095 𝑚 𝑉𝑚𝑖𝑛 is determined by the available power and > 𝑉𝑠𝑡𝑎 𝑙𝑙 !

Example: Consider a jet A/C with

𝑏 = 16.25 𝑚

𝑆 = 29.54 𝑚2

𝑊 = 88251 𝑁

𝑇𝑚𝑎𝑥 = 16250 𝑁 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙

𝐶𝐷0= 0.02

𝑒 = 0.81

Find the maximum speed of the A/C at 6000 𝑚.

𝑇𝐴 𝑚𝑎𝑥 @6000𝑚 =𝜌

𝜌0

𝑇𝐴 𝑚𝑎𝑥 @0𝑚

= 0.5389× 16250

= 8757 𝑁

𝑉𝑚𝑎𝑥 can then be found from the (𝑉∞ eqn) as

𝑉𝑚𝑎𝑥 =

𝑇𝐴 𝑚𝑎𝑥

𝑆 ±𝑊𝑆

𝑇𝐴 𝑚𝑎𝑥𝑊

2

− 4𝐶𝐷0

1

𝜋𝑒𝑏2

𝑆

𝜌𝐶𝐷0

1 2

= 201𝑚

𝑠

= 0.64 𝑀

Note that we ignored wave drag due to air compressibility, which could actually be significant at this speed!

Takeoff Performance

Ground Roll

0 50 100 150 200 2500.5

1

1.5x 10

4

V (m/s)

TR (

N)

0 50 100 150 200 2500

0.5

1

1.5

2

2.5

3

3.5x 10

6

V (m/s)

PR

(W

)


23

Thrust, 𝑇

Drag, 𝐷 =1

2𝜌∞𝑉∞

2𝑆𝐶𝐷

Lift, 𝐿 =1

2𝜌∞𝑉∞

2𝑆𝐶𝐿

Weight, 𝑊 = 𝑚𝑔

Resistance Force, 𝑅 = 𝜇𝑟 𝑊 − 𝐿 , 𝜇𝑟 : Coefficient of rolling friction

Balance of forces parallel to ground:

𝐹∥ = 𝑇 − 𝐷 − 𝑅 = 𝑚𝑑𝑉∞𝑑𝑡

When rolling on ground

𝐶𝐷 = 𝐶𝐷0 + 𝜙𝐶𝐿

2

𝜋𝑒𝐴𝑅

where 𝜙 is a coefficient signifying the ground effects. When an A/C flies close to ground the effects due to

wing-tip vortices are smaller. Near ground the induced drag is not very strong.

𝜙 = 16𝑕 𝑏 2

1 + 16𝑕 𝑏 2

where 𝑕 is the height of the wing above the ground and 𝑏 is the wingspan.

𝐷, 𝑅 (because of 𝐿), and possibly 𝑇 are all functions of 𝑉∞ . We can get an approximate solution by using

the average of the net force during ground roll:

𝐹𝑒𝑓𝑓 = 𝑇 − 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒 = 𝑚𝑑𝑉∞𝑑𝑡

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Then

𝐹𝑒𝑓𝑓 = 𝑚𝑑𝑉∞𝑑𝑡

Integrating

𝑑𝑉∞

𝑉𝐿𝑂

0

= 𝐹𝑒𝑓𝑓

𝑚𝑑𝑡

𝑡𝐿𝑂

0

𝑉𝐿𝑂 =𝐹𝑒𝑓𝑓

𝑚𝑡𝐿𝑂

𝑡𝐿𝑂 =𝑉𝐿𝑂𝑚

𝐹𝑒𝑓𝑓

𝑡𝐿𝑂: time to accelerate mass 𝑚 to velocity 𝑉𝐿𝑂 under a fixed force 𝐹𝑒𝑓𝑓 .

𝑥 = 0

𝑡 = 0

𝑉∞ = 0

𝑥 𝑥 = 𝑥𝐿𝑂

𝑡 = 𝑡𝐿𝑂

𝑉∞ = 𝑉𝐿𝑂

𝑚 𝐹𝑒𝑓𝑓 𝑚 𝐹𝑒𝑓𝑓


24

Distance covered:

𝑉 =𝑑𝑥

𝑑𝑡

𝑑𝑥 = 𝑉𝑑𝑡

𝑑𝑥

𝑥𝐿𝑂

0

=𝐹𝑒𝑓𝑓

𝑚 𝑡𝑑𝑡

𝑡𝐿𝑂

0

𝑥𝐿𝑂 = 𝐹𝑒𝑓𝑓

𝑚

𝑡𝐿𝑂2

2

=𝑉𝐿𝑂

2 𝑚

2𝐹𝑒𝑓𝑓

𝑥𝐿𝑂: distance to accelerate mass 𝑚 to velocity 𝑉𝐿𝑂 under a fixed force 𝐹𝑒𝑓𝑓 .

Remember

𝐹𝑒𝑓𝑓 = 𝑇 − 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒

𝑥𝐿𝑂 =𝑉𝐿𝑂

2 𝑚

2 𝑇 − 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒

Liftoff velocity 𝑉𝐿𝑂 depends on the aerodynamic design. It is typically taken as

𝑉𝐿𝑂 = 1.2𝑉𝑠𝑡𝑎𝑙 𝑙 = 1.2 2𝑊

𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥

Then

𝑥𝐿𝑂 =1.44𝑊2

𝑔𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥 𝑇 − 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒

If

𝑇 ≫ 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒

𝑥𝐿𝑂 ≅1.44𝑊2

𝑔𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥 𝑇

Usually we can calculate

𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒 = 𝐷 + 𝜇𝑟 𝑊 − 𝐿 𝑉∞=0.7𝑉𝐿𝑂

where 𝑉𝐿𝑂 = 1.2𝑉𝑠𝑡𝑎𝑙𝑙 .

𝑥𝐿𝑂 equation illustrates that:

- Liftoff distance is very sensitive to the weight of the A/C, varying directly as 𝑊2.

- Liftoff distance depends on the air density 𝜌∞ . If we assume that thrust is directly proportional to

𝜌∞ , then

𝑥𝐿𝑂 ∝1

𝜌∞2

This is why on hot summer days an A/C requires a longer ground roll.

- The liftoff distance can be decreased by increasing the wing area, increasing 𝐶𝐿𝑚𝑎𝑥 , and increasing

thrust, all of which make sense.


25

Example: Consider our sample piston A/C:

𝑏 = 10.91 𝑚

𝑆 = 16.17 𝑚2

𝑊 = 13139 𝑁

𝑇𝑚𝑎𝑥 = 7000 𝑁 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 𝑎𝑛𝑑 𝑙𝑜𝑤 𝑠𝑝𝑒𝑒𝑑

For a clean wing configuration 𝐶𝐿𝑚𝑎 𝑥 = 1.5 and the drag polar is given by

𝐶𝐷 = 𝐶𝐷0+ 𝐾𝐶𝐿

2

with 𝐶𝐷0= 0.025 and 𝐾 = 0.05. Assume that during ground roll maximum thrust is fixed at 𝑇𝐴 𝑚𝑎𝑥 =

7000 𝑁. Find 𝑥𝐿𝑂 at sea level.



𝑚

𝑠

𝐹𝑒𝑓𝑓 = 𝑇 − 𝐷 + 𝜇𝑟 𝑊 − 𝐿 0.7×1.2𝑉𝑠𝑡𝑎𝑙𝑙

= 6152 𝑁

𝑥𝐿𝑂 =1.44𝑊2

𝑔𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥 𝐹𝑒𝑓𝑓

= 138.6 𝑚

Consider the same A/C with flaps extended halfway. For this configuration the A/C has a new drag polar

with

𝐶𝐷0= 0.03

𝐾 = 0.1

Then we have

𝑉𝑠𝑡𝑎𝑙𝑙 = 25.8𝑚

𝑠

𝐹𝑒𝑓𝑓 = 5334 𝑁

𝑥𝐿𝑂 = 119.9 𝑚

If the flaps are extended all the way, the drag polar coefficients change to:

𝐶𝐷0= 0.035

𝐾 = 0.15

Then we have

𝑉𝑠𝑡𝑎𝑙 𝑙 = 23.0𝑚

𝑠

𝐹𝑒𝑓𝑓 = 4177 𝑁

𝑥𝐿𝑂 = 122.5 𝑚

The liftoff distance is greater! That’s why flaps are usually not fully extended for takeoff!


26

Rate of Climb How fast can an A/C climb?

Consider an A/C in steady unaccelerated climbing flight with 𝛼 ≪ 𝛾

𝑤: Rate of climb (𝑅/𝐶), vertical component of the flight velocity

𝑢: Horizontal component of the flight velocity

Sum of forces parallel to flight path is equal to zero (unaccelerated flight!):

𝑇 = 𝐷 + 𝑊 sin 𝛾 (∗)

and perpendicular to flight path

𝐿 = 𝑊 cos 𝛾 𝐿 < 𝑊 !

Multiply (*) by 𝑉∞ and divide by 𝑊:

𝑇 − 𝐷 𝑉∞𝑊

= 𝑉∞ sin𝛾 = 𝑤 = 𝑅/𝐶

𝑇 − 𝐷 𝑉∞ = Excess Power

𝑇𝐴 − 𝑇𝑅 𝑉∞ = Excess Power

So,

𝑅/𝐶 =Excess Power

𝑊

Note: Normally the 𝑃𝑅 vs 𝑉∞ curves are generated for level flight conditions and

𝑃𝑅 @ level flight ≠ 𝑃𝑅@ climbing flight

P (

W)

P = TV1

PR = T

RV

1 = DV

1

= (T - D)V1

PR

V1

V

(m/s)

Excess

power

Propeller A/C

P (

W)

PA

PR

V

(m/s)

Excesspower

Jet A/C

𝛾

𝐷

𝐿

𝑇

𝑊 𝛾

𝑉∞

𝑢 = 𝑉∞ cos 𝛾

𝑉∞ 𝑤 = 𝑉∞ sin 𝛾 𝛾


27

But for small 𝛾 we may assume that they are equal and use 𝑃𝑅 vs 𝑉∞ curve generated for level flight to find

the excess power.

𝐿 @ climbing flight < 𝐿 @ level flight for same 𝑉∞

𝐷 @ climbing flight < 𝐷 @ level flight for same 𝑉∞

max𝑅/𝐶 =maximum Excess Power

𝑊

An important difference in the low-speed R/C performance can be seen between a propeller driven and a

jet A/C. Due to the 𝑃𝐴 characteristics of a piston engine-propeller combination, large excess powers are

available at low values of 𝑉∞ just above the stall. In contrast the excess power available to jet A/C at low 𝑉∞

is small, with a correspondingly reduced R/C capability.

Time to Climb The R/C was defined as

𝑅 𝐶 = 𝑤 =𝑑𝑕

𝑑𝑡

P (

W)

(PA)max

= (TA)max

V

(PR)min

= (TR)min

V

= Dmin

V

= ((TA)max

- (TR)min

)V

PR

Vmax R/C

Vmax

V

(m/s)

Max

excess

power

Propeller A/C

R/C

(m

/s)

Vmax R/C

max R/C

Vmax

V

(m/s)

0

Determination of 𝑚𝑎𝑥 𝑅/𝐶 for a given altitude graphically.


28

Then

𝑑𝑡 =𝑑𝑕

𝑅 𝐶

Time to climb from altitude 𝑕1 to 𝑕2 can be found by integrating as

𝑡𝑐 = 𝑑𝑡

𝑡𝑐

0

= 𝑑𝑕

𝑅 𝐶

𝑕2

𝑕1

Note that 𝑅 𝐶 changes with altitude at a fixed 𝑉∞ . Both 𝑃𝑅 and 𝑃𝐴 (and hence excess power) change with

altitude.

To compute 𝑡𝑐 graphically we need to plot 𝑅 𝐶 −1 versus 𝑕 first. The 𝑡𝑐 for our sample piston A/C for

climbing from 2000 𝑚 to 6000 𝑚 at 𝑉∞ = 60 𝑚/𝑠 can be computed from the above figure on the right.

The area under the 𝑅 𝐶 −1 versus 𝑕 curve gives 𝑡𝑐 .

Absolute Ceiling and Service Ceiling Max excess power available decreases with increasing altitude. There exists an altitude at which the max

excess power available, and hence R/C becomes zero. This altitude is defined as the absolute ceiling.

A more useful quantity is service ceiling, defined as

the altitude where the 𝑅 𝐶 𝑚𝑎𝑥 = 100 𝑓𝑡/𝑚𝑖𝑛.

Absolute and service ceilings can be determined as

follows:

1. Calculate values of 𝑅 𝐶 𝑚𝑎𝑥 for a number

of different altitudes.

2. Plot altitude versus 𝑅 𝐶 𝑚𝑎𝑥

3. Find the absolute and service ceilings by

interpolation at 𝑅 𝐶 𝑚𝑎𝑥 = 0 and

𝑅 𝐶 𝑚𝑎𝑥 = 100 𝑓𝑡/𝑚𝑖𝑛 respectively.

For our sample piston A/C the ceilings are found on

the figure to the right.

P

PA1

PR1

PA2

PR2

V

Propeller A/C

Excess

power 1

Excess

power 2

0 2000 4000 6000 80000

0.1

0.2

0.3

0.4

0.5

Area = tc

h (m)

(R/C

)-1 (

s/m

)

0 500 1000 15000

2000

4000

6000

8000 Absolute ceiling (~8177 m)

Service ceiling (~7527 m)

Not linear!

(R/C)max

(ft/min)

h (

m)


29

Example: Consider the piston engine A/C of HW#2. Find the maximum R/C at the altitude where

𝜌𝑕 = 0.6 𝑘𝑔/𝑚3.

The maximum R/C is achieved at 𝑃𝑅 𝑚𝑖𝑛 and at 𝐶𝐷𝑖 = 3𝐶𝐷0. So,

𝐶𝐿2

𝜋𝑒𝐴𝑅= 3𝐶𝐷0

⇒ 𝐶𝐿 = 3𝐶𝐷0𝜋𝑒𝐴𝑅

The speed at which 𝑃𝑅 𝑚𝑖𝑛 occurs can be found from

𝑊 = 𝐿 =1

2𝜌𝑕𝑉 𝑃𝑅 𝑚𝑖 𝑛

2 𝑆 3𝐶𝐷0𝜋𝑒𝐴𝑅 ⇒ 𝑉 𝑃𝑅 𝑚𝑖𝑛

2 =2𝑊

𝜌𝑕𝑆 3𝐶𝐷0𝜋𝑒𝐴𝑅

𝑃𝑅 𝑚𝑖𝑛 can then be found from

𝑃𝑅 𝑚𝑖𝑛 = 𝑇𝑅 𝑚𝑖𝑛 𝑉 𝑃𝑅 𝑚𝑖𝑛

=1

2𝜌𝑕𝑉 𝑃𝑅 𝑚𝑖𝑛

3 𝑆 𝐶𝐷0+ 𝐶𝐷𝑖

=3𝐶𝐷0

𝑃𝐴 𝑚𝑎𝑥 at the given altitude is given by

𝑃𝐴 𝑚𝑎𝑥 ,𝑕 =𝜌𝑕𝜌0𝜂𝑃𝑟𝑃𝑚𝑎𝑥 ,0

The excess power is given by

𝐸𝑃 = 𝑃𝐴 𝑚𝑎𝑥 ,𝑕 − 𝑃𝑅 𝑚𝑖𝑛

=𝜌𝑕𝜌0𝜂𝑃𝑟𝑃𝑚𝑎𝑥 ,0 − 2𝜌𝑕𝑉 𝑃𝑅 𝑚𝑖𝑛

3 𝑆𝐶𝐷0

𝐸𝑃 𝑕 =𝜂𝑃𝑟𝑃𝑚𝑎𝑥 ,0

𝜌0𝜌𝑕 −

2𝑆𝐶𝐷0

𝜌𝑕

2𝑊

𝑆 3𝐶𝐷0𝜋𝑒𝐴𝑅

3 2

(𝐸𝑃 − 𝑕 eqn)

Putting in the numbers with 𝜌𝑕 = 0.6 𝑘𝑔/𝑚3 we find

𝐸𝑃 = 53487 𝑊𝑎𝑡𝑡

𝑅/𝐶 𝑚𝑎𝑥 @𝑕 =𝐸𝑃

𝑊= 1.05 𝑚/𝑠 = 206 𝑓𝑡/𝑚𝑖𝑛

Find the absolute and service ceilings.

At absolute ceiling 𝐸𝑃 𝑕 = 0, so we are looking for 𝑕𝑎𝑏𝑠 .𝑐 . (or equivalently 𝜌𝑕𝑎𝑏𝑠 .𝑐 .) for which 𝐸𝑃 𝑕𝑎𝑏𝑠 .𝑐 . =

0. By equating the left hand side of the (𝐸𝑃 − 𝑕 eqn) to zero we get

𝜌𝑕𝑎𝑏𝑠 .𝑐 .=

2𝜌0𝑆𝐶𝐷0

𝜂𝑃𝑟𝑃𝑚𝑎𝑥 ,0

2 3

2𝑊


= 0.5152 𝑘𝑔/𝑚3

From the ISA table by interpolation we can find

𝑕𝑎𝑏𝑠 .𝑐 . = 8177 𝑚

At service ceiling

𝑅/𝐶 𝑚𝑎𝑥 =𝐸𝑃𝑚𝑎𝑥𝑊

= 100 𝑓𝑡/𝑚𝑖𝑛 = 0.508 𝑚/𝑠


30

We need to solve 𝐸𝑃 𝑕𝑠𝑒𝑟 .𝑐 . = 0.508𝑊

𝜂𝑃𝑟𝑃𝑚𝑎𝑥 ,0

𝜌0𝑥3 − 0.508𝑊𝑥 − 2𝑆𝐶𝐷0

2𝑊


3 2

= 0

where 𝑥 = 𝜌𝑕𝑠𝑒𝑟 .𝑐 .. This gives

𝜌𝑕𝑠𝑒𝑟 .𝑐 .= 0.5556 𝑚/𝑘𝑔3

that corresponds to

𝑕𝑠𝑒𝑟 .𝑐 . = 7526 𝑚

Accelerated Rate of Climb Airplanes usually perform accelerated climb. Exact expressions can be derived using equations of motion,

but instead we will use the energy method for convenience.

𝑇𝑜𝑡𝑎𝑙 𝐴/𝐶 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑃𝐸 + 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝐾𝐸

𝐸 = 𝑚𝑔𝑕 +1

2𝑚𝑉2

Define specific energy as

𝐻𝑒 =𝑃𝐸 + 𝐾𝐸

𝑊=𝑚𝑔𝑕 +

12𝑚𝑉2

𝑚𝑔= 𝑕 +

𝑉2

2𝑔

𝐻𝑒 has the unit of height and is therefore also called the “energy height” of the A/C.

How does an A/C change its energy state? Let’s look at the equations of motion:

𝑇 − 𝐷 −𝑊 sin𝛾 = 𝑚𝑑𝑉

𝑑𝑡=𝑊

𝑔

𝑑𝑉

𝑑𝑡

𝑇𝑉 − 𝐷𝑉

𝑊= 𝑉 sin𝛾 +

𝑉

𝑔

𝑑𝑉

𝑑𝑡

Recall that

0 50 100 1500

500

1000

1500

2000

V

(m/s)

h (

m)

Constant energy height contours

He = 2000 m

He = 1500 m

He = 1000 m

He = 500 m 𝛾

𝐷

𝐿

𝑇

𝑊

𝛾


31

𝑉 sin 𝛾 = 𝑅/𝐶 =𝑑𝑕

𝑑𝑡

𝑇𝑉 − 𝐷𝑉

𝑊=𝐸𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟

𝑊= 𝐸𝑃𝑠 = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑒𝑥𝑐𝑒𝑠𝑠 𝑝𝑜𝑤𝑒𝑟

Then,

𝐸𝑃𝑠 =𝑑𝑕

𝑑𝑡+𝑉

𝑔

𝑑𝑉

𝑑𝑡

This equation states that an A/C can use the available excess power for rate of climb (𝑑𝑕/𝑑𝑡) or to

accelerate along its flight path (𝑑𝑉/𝑑𝑡), or for combination of both.

Return to the energy height equation

𝐻𝑒 = 𝑕 +𝑉2

2𝑔

Differentiating with respect to time we have

𝑑𝐻𝑒𝑑𝑡

=𝑑𝑕

𝑑𝑡+𝑣

𝑔

𝑑𝑉

𝑑𝑡= 𝐸𝑃𝑠

The time rate of change of energy height is equal to the specific excess power. This implies that an A/C can

increase its energy state simply by application of excess power.

Gliding Flight In a gliding flight, power is off (𝑇 = 0).

Equation of motion parallel to flight path:

𝑊 sin 𝛾 − 𝐷 = 0

Equation of motion perpendicular to flight path:

𝐿 −𝑊 cos𝛾 = 0

Eliminating 𝑊 we can write the gliding angle as

tan 𝛾 =1

𝐿/𝐷

For smallest glide angle

tan 𝛾 𝑚𝑖𝑛 =1

𝐿/𝐷 𝑚𝑎𝑥

𝛾 𝐷

𝐿

𝑇

𝑊 𝛾 𝑉∞


32

For unpowered flight 𝑇 = 0 and hence

𝐸𝑃𝑠 =𝑇𝑉 =0

−𝐷𝑉

𝑊< 0

So

𝐸𝑃𝑠 =𝑑𝑕

𝑑𝑡+𝑉

𝑔

𝑑𝑉

𝑑𝑡< 0

In unpowered flight an A/C loses altitude, or speed, or both! For a gliding flight in equilibrium we assume

𝑑𝑉/𝑑𝑡 = 0, other wise the flight cannot stay in equilibrium for long. So we have

𝑑𝑕

𝑑𝑡=−𝐷𝑉

𝑊

𝑉 sin𝛾 =−𝐷𝑉

𝐿cos𝛾

tan 𝛾 = −1

𝐿 𝐷

If we take 𝛾 below horizontal to be positive

tan 𝛾 =1

𝐿 𝐷

We get the same glide angle equation using the energy method.

𝑅 =𝑕

tan 𝛾

So for maximum range

𝑅𝑚𝑎𝑥 =𝑕

tan 𝛾 𝑚𝑖𝑛= 𝑕 𝐿/𝐷 𝑚𝑎𝑥

Recall that

𝐿

𝐷=𝐶𝐿𝐶𝐷

=𝐶𝐿

𝐶𝐷0+ 𝐾𝐶𝐿

2

𝛾

𝐷

𝐿

𝑇

𝑊 𝛾 𝑉∞

𝛾

𝑕

𝑅


33

does not depend on altitude, but depends on 𝐶𝐿, and hence 𝛼. So for a particular 𝐿 𝐷 the A/C needs to fly

at a particular 𝐶𝐿, and this dictates the equilibrium glide velocity.

𝐿 = 𝑊 cos 𝛾 =1

2𝜌∞𝑉∞

2𝑆𝐶𝐿

𝑉∞ = 2 cos 𝛾

𝜌∞𝐶𝐿

𝑊

𝑆

Equilibrium glide velocity depends on altitude through 𝜌∞ and wing loading. If 𝐿 𝐷 is held constant

throughout the glide path, then 𝐶𝐿 is constant along the glide path. However, the equilibrium velocity will

change with altitude, decreasing with decreasing altitude.

Landing Performance Consider an A/C during landing. Typically the thrust is reduced to zero at touchdown. Therefore the

equation for landing ground roll is obtained from the takeoff ground roll equation by setting 𝑇 = 0.

−𝐷 − 𝑅 = −𝐷 − 𝜇𝑟 𝑊 − 𝐿 = 𝑚𝑑𝑉∞𝑑𝑡

Assume

𝐹𝑒𝑓𝑓 = −𝐷 − 𝜇𝑟 𝑊 − 𝐿 𝑎𝑣𝑒 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

= −𝐷 − 𝜇𝑟 𝑊 − 𝐿 𝑉=0.7𝑉𝑇𝐷

where 𝑉𝑇𝐷 is the touchdown velocity and can be taken

as 𝑉𝑇𝐷 = 1.3𝑉𝑠𝑡𝑎𝑙𝑙 . In most cases it is very reasonable

to approximate −𝐷 − 𝜇𝑟 𝑊 − 𝐿 with a constant

force.

𝑥𝐿𝑎𝑛𝑑 =1.69𝑊2

𝑔𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥 𝐷 + 𝜇𝑟 𝑊 − 𝐿 0.7𝑉𝑇𝐷

Thrust reversing can also be used to reduce the

distance required to stop the A/C.

𝑥𝐿𝑎𝑛𝑑 =1.69𝑊2

𝑔𝜌∞𝑆𝐶𝐿𝑚𝑎𝑥 𝑇𝑅 + 𝐷 + 𝜇𝑟 𝑊 − 𝐿 0.7𝑉𝑇𝐷

where 𝑇𝑅 is the reversed thrust.

0

0

V

VTD

0.7VTD

Real Feff

(Feff

)ave

D

R

- (D + R)

ae172lecturenoteskutay

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