conception aero performances

8/12/2019 Conception Aero Performances

1/96

Introduction to Aircraft Design

Aircraft Design

Lecture 8:

Aircraft Performance

G. Dimitriadis


2/96


Design for Performance

The most important requirement for a newaircraft design is that it fulfills its mission

This is assured through performancecalculations at the design stage

As these calculations are carried out,important aircraft parameters are chosen: Size of wing

Engine


3/96


Flight points

Performance calculations are crucial atseveral flight points. The most importantare:

Cruise

Take off

Climb

Landing

The first performance design analyses foran airliner are carried out for cruise


4/96


Weight and drag

In order to choose parameters such as

engine and wing size, the aircrafts

weight and drag must be known.

Then the amount of lift and thrust

required can be determined

Of course, the weight and drag must be

calculated at several important points inthe flight envelope.


5/96


Methodology

It is impossible to know the weight and

drag of an aircraft before it has even

been designed There are two possibilities:

Carry out detailed simulations at the

conceptual design stage; very costly

Use previous experience, statistical data,carry out industrial espionage etc


6/96


Statistics

There is an enormouswealth of data on civil

aircraft. There are

hundreds of types thatare very similar. They can

be used to extract somevery useful statistics.


7/96


Weight guesstimates

The first important weight to calculate is thetake off weight, Wto.

It is usually expressed as:

Where Wpis the payload weight, Wfis the fuelweight, Wfixis the fixed empty weight (e.g.engine) and Wvaris the variable empty weight.

Notice that in this expression, We=Wfix+Wvaristhe total empty weight of the aircraft

Wto=

Wp +Wfix

1-Wvar

Wto

Wf

Wto


8/96


Light aircraft (Wto


9/96


Undercarriage drag correction

factor The undercarriage drag correction factor is used

both in the calculation of fuel weight and in thecalculation of the zero-lift drag (see later)

For a fully retractable landing gear that disappears

inside the aircraft lines, ruc=1. Otherwise:

ruc =

1.35 - for fixed gear without streamlined wheel fairings

1.25 - for fixed gear with streamlined wheel fairings

1.08 - main gear retracted in streamline fairings on the fuselage

1.03 - main gear retracted in engine nacelles

"

#

$$

%$$


10/96


Wheel Fairings

Cessna 180 with wheelfairing

Cessna 172 without wheelfairing


11/96


Transport Aircraft (Wto>5670Kg) Again, statistical studies show that

Where Wengis the engine weight and Weis calculated from a statistical graph. Allweights are in Kg.

The fuel weight is also calculated from astatistical graph

Wvar

Wto

= 0.2

Wfix =Weng + 500+ We


12/96


We

calculation

lf = length of fuselagebf= width of fuselage

hf = height of fuselage

Use metric data


13/96


Fuel weight -

Turboprops

Cp= Specific fuel

consumption forpropeller aircraft

Use metric data


14/96


Fuel weight -

Turbojetsp : atmospheric pressure at cruiseconditions

M : Mach number at cruiseconditions

=T/T0, cruise temperature/

standard temperatureCT/: Corrected specific fuel

consumption at cruise conditionsa0: Speed of sound at sea level,

International Standard

Atmosphere: Mean skin friction coefficient

based on wetted area.

CF=

0.003 - for large, long- range transports

0.0035 - for small, short - range transports

0.004 - for business and executive jets

"

#$

%$

CF


15/96


Skin friction coefficient

An estimated of the drag force due to air frictionover the full surface of the aircraft (wettedarea).

It can be estimated using Prandtl-Schlichtingtheory as

WhereRecris the Reynolds number based oncruise flight conditions and the fuselage length

CF=

0.455

log10 Recr( )( )2.58


16/96


More skin friction

Skin frictioncoefficients for

several aircraft

types.

rRe=C

F Re( ) /CF 10

8( )


17/96


Caravelle type


18/96


Boeing 707 type


19/96


Boeing 727 type


20/96


Avro RJ85/100 (Bae 146)


21/96


Antonov An-72/74


22/96


Turboprops


23/96


Blended Wing Body


24/96


Box wing


25/96


Circular wing


26/96


Drag Calculations

There are many sources of aircraft drag

They are usually summarized by the

airplane drag polar:

Where CD0is the drag that is

independent of lift and eis the Oswaldefficiency factor.

CD= C

D0

+

CL

2

eAR


27/96


The drag polar

At high values of CLthe wing stalls


28/96


Drag figures for different aircraft

Aircraft Type CD0 e

High-subsonic jet 0.014-0.020 0.75-0.85

Large turboprop 0.018-0.024 0.80-0.85

Twin-engine pistonaircraft

0.022-0.028 0.75-0.80

Single-engine pistonaircraft with fixed gear

0.020-0.030 0.75-0.80

Single-engine pistonaircraft with retractable

gear

0.025-0.040 0.65-0.75

Agricultural aircraftwithout spray system

0.060 0.65-0.75

Agricultural aircraft withspray system

0.070-0.080 0.65-0.75


29/96


Compressibility drag

Compressibility effects increase drag

A simple way of including

compressibility effects inpreliminary drag calculations is

to add CDto CD0, where

CD=0.0005 for long-range

cruise conditions

CD=0.0020 for high-speedcruise conditions


30/96


Cruise


31/96


High speed performance

At cruise, the flight speed is constant.Therefore, T=D. This can be written as

Where is the cruise air density, Vis the cruise

airspeed and Sis the wing area. The drag coefficient is obtained from the drag

polar, using the fact thatL=W, i.e.

T= D =1

2V

2CDS

W = L =1

2V

2CLS

CL= W1

2V

2S


32/96


Thrust to weight

The thrust to weight ratio is then

The thrust here is the installed thrust,

which is 4-8% lower than the un-

installed thrust

This equation can be used to choose anengine for the cruise condition

T

W=

1

2WV

2CD

0

+

CL

2

eAR

$

%&

'

()S=

V2CD

0

2W /S+

2W

eARV2S


33/96


Minimum Thrust

The thrust-to-weight ratio can be minimized asa function of wing loading W/S.

Minimum thrust is required when

Where d1=0.008-0.010 for aircraft withretractable undercarriage. Therefore

W

S=

1

2V

2d1eAR

T

W"#$ %

&'min

= CD0 + d1

d1eAR


34/96


Engine thrust

The thrust of an engine at the cruisecondition can be determined from:

Manufacturers data

Approximate relationship to the take offthrust:

where is the bypass ratio,Mis the cruise

Mach number and G=0.9 for low bypassengines and G=1.1 for high bypass

T

Tto

= 10.454 1+ ( )

1+ 0.75M+ 0.6 +

0.13

G

$%

&'M

2


35/96


Wing loading and Mach

numberThe maximum wing loadingdepends uniquely on CLmax.

For an airliner, CLmaxdrops

to neary zero asM

approaches 1.

For supersonic aircraft,

CLmaxdrops in the transonic

region but not too much. It

then assumes a near

constant value for moderatesupersonic speeds


36/96


The flight envelopeMMO,VMO=maximumoperating Mach number,

airspeed

MC, VC=design cruising

Mach number, airspeedMD,VD=design diving Machnumber, airspeed

VS=stalling airspeed

CAS=calibrated airspeed

(airspeed displayed on flight

instruments)TAS=true airspeed


37/96


Determination of cruise

Mach number

MHS=high-speedMach number (for

high-speed cruise)

This calculation must

be repeated atseveral altitudes.

Each altitude

corresponds to a

different cruise Mach.


38/96


Range performance

The range of an aircraft can be estimated

from the Brguet range equation:

This equation is applicable to cruise

conditions only, i.e.L/Dis the cruise lift-to-

drag ratio, Vis the cruise airspeed, Wiis

the weight of the aircraft at the start ofcruise and Wf is the cruise fuel weight

R =V

CT

L

D

ln W

i

WiWf

#

$%

&

'(


39/96


Maximizing Range

The range equation can also be written as

WhereMis the cruise Mach number anda0is the speed of sound at sea level.

The range can be maximized either by

maximizingL/Dor by maximizingML/D.

Therefore:

R

a0

=

ML /D

CT /

ln W

i

WiW

f

$

%&

'

()

CL= C

D0eAR or CL = 1

3CD0eAR


40/96


Lift to drag ratio

Example of lift to dragratio variation with lift,

angle of attack and the

deployment of slats and

flaps

The best L/D is obtained

for the clean configuration

Induced drag only


41/96


Compressibility effect on

rangeWith compressibility thereis a global maximumML/D

Without compressibility there isno global maximum


42/96


43/96


Some considerations

For long-haul aircraft the most importantconsideration is fuel efficiency

For short-haul aircraft the most importantconsideration is engine weight

There are some complications though: Cruise fuel is only part of the fuel weight

For short haul aircraft the engine thrust isfrequently determined by take off field length

Air traffic controllers decide the allowable cruise

altitudes

An aircraft can have more than one engine


44/96


Reserve Fuel One definition of reserve fuel for international

airline operations (ATA 67): The airliner must carry enough reserve fuel to:

Continue flight for time equal to 10% of basic flighttime at normal cruise conditions

Execute missed approach and climbout at

destination airport Fly to alternate airport 370km distant

Hold at alternate airport for 30 minutes at 457mabove the ground

Descend and land at alternate airport

One approximate calculation:W

fres/W

to=0.18C

T / AR


45/96


Range for propeller aircraft

The Brguet range equation for propelleraircraft is

Where pis the propeller efficiency and Cpis the specific fuel consumption.

The range can be maximized by: Minimizing airplane drag

Minimizing engine power

R =p

CP

L

Dln

Wi

WiW

f

$

%&

'

()


46/96


Payload-range diagram

Two cases: high-speed cruise, long-rangecruise. Example: A-300B


47/96


Take off


48/96


Take Off Performance

T

D

L

R

W

V

x

V1VLOFV=0

V2

h=35ft

Ground Run Rotation Transition Climb

xg xa


49/96


Take Off Description

Take off starts at time t0, with airspeed V0andthe runway may have an angle to horizontalof .

Lift off occurs at time tg, after a distance ofxg,usually at speed VLOF.

Take off is completed when the aircraft hasreached sufficient height to clear an obstacle35ft high (50ft for military aircraft)

Finally, the climb out phase takes the aircraftto 500ft at the climb throttle setting.


50/96


Ground Run

The ground run can either start at V0=0.

The angle of attack is defined with respect tothe thrust line. For aircraft with a tricycle landing gear the angle of

attack is nearly zero. At a sufficient speed thenose is lifted up a little bit to achieve an optimumangle of attack.

For aircraft with a tailwheel the angle of attack isvery high. At an airspeed where the controlsurfaces are effective the angle of attack is

decreased to decrease the drag and increaseacceleration.


51/96


Lift off

In principle an aircraft can lift off once it hasreached its stall speed, Vstall.

At this airspeed it can increase its angle of

attack max, with a resulting CLmax. This liftcoefficient will generally satisfyL>W.

However, for safety reasons, the lift off speedis defined as VLOF=k1Vstallwhere

k1=1.10is an indicative value

In general k1varies with type of aircraft


52/96


Rotation and Transition

After lift off the aircraft rotates to deflect thevelocity from nearly horizontal to a fewdegrees upward.

The rotation lasts usually for a couple of

seconds. The transition stage follows, at the end of

which the aircraft is required to clear ahypothetical obstacle.

The speed at the clearance of the obstacle isusually V2=k2Vstallwhere k2=1.2.


53/96


Take off details

V1is called the decision speed for multi-engined aircraft. Below this speed, if one engine fails, the take off is

aborted.

Above this speed, if one engine fails, the take off

is continued. Take off field length: Distance from rest to

obstacle clearance

The climb stage follows the transition stage.

The take off phase is usually considered to be

complete at around 500ft once the flaps havebeen retracted.


54/96


Equations of motion

The total force applied to the aircraft parallelto the runway is:

The acceleration is equal to a=dV/dt. Considering that the distancexis given by dx/

dt=V,

dx

dt

dt

dV=

V

a so that

xg=

V

a0

VLOF

dV

F = TR D = ma


55/96


Friction force

In order to have a friction force, it must

be thatR>0, i.e.

Where it is assumed that the runways

inclination is small. The total force can

be written as

R =W L =W1

2 V

2SC

L

ma = TW 1

2V

2S C

DC

L( )


56/96


Friction Coefficients

Runway type

Concrete, asphalt 0.02

Hard turf 0.04

Field with short grass 0.05

Field with long grass 0.10

Soft field, sand 0.10-0.30


57/96


Approximate solution (1)

An approximate solution for the ground

run can be obtained as

xg V

LOF

2/2g

TW

to

$

T = thrust at VLOF

/ 2 0.755 +

4 + T

to

CL = eAR

$ = + 0.72CD0

CLmax

where


58/96


Approximate solution (2)

An approximate solution for the air run

can be obtained as

The airspeed at the takeoff obstacle is

xa V

LOF

2

g 2+

h

LOF

where

LOF

=

TD

W

$

%&

'

()LOF

0.9T

Wto

0.3

AR

V2=V

LOF 1+

LOF 2


59/96


60/96


Climb


61/96


Climb performance

Operational requirements, e.g.

Rate of climb at sea level

Service ceiling altitude for a maximum rate

of climb of 0.5m/sAirworthiness requirements

Minimum climb gradient in takeoff, en

route, landing

Rate if climb at a specified altitude with oneengine inoperative


62/96


Climb description

Climb follows immediately after take off andits objective is to bring the aircraft to cruisingheight

In general, climb is performed in a vertical

plane, i.e. no turning during climb.

Climb takes up a relatively short percentageof flight time so it can be assumed that theaircrafts weight remains constant

Variations in speed and flight path areconsidered small, i.e. constant speed climb


63/96


Climb diagram

L

D

VT

W

x

z

Notice that thethrust line is not

necessarily

parallel to the flight

path

Also note that, for

constant speed V,

the aircraft cannot

be accelerating


64/96


Thrust and Lift equations

Assume that =0. Then, write theequilibrium equations on the body axes:

As mentioned before, the drag can be

expressed in terms of lift using the dragpolar

T =1

2

SV2CD +Wsin

Wcos =1

2SV

2CL


65/96


Thrust and Power

It is customary to express the thrust equationin terms of power by multiplying it by V:

Define

WherePais the power available for climb,Pris the power required for level flight at thesame airspeed and Vzis the rate of climb.

Then

TV =1

2SV

3CD +WVsin

Pa = TV, P

r =

12SV

3C

D, V

z = Vsin

Vz =

Pa

Pr

W and sin =

Pa

Pr

VW


66/96


Climb of a jet aircraft

For jet aircraft the amount of power available

for climb varies linearly with airspeed.

The power required for level flight varies non-

linearly with airspeed.

There is an optimum airspeed for maximum

rate of climb which maximizesPa-Pr.

This airspeed is not necessarily at the

airspeed yielding the minimum value ofPr.


67/96


Optimum airspeedThe optimum climb rate isobtained when the powerdifference Pismaximized.

It occurs at the pointwhere the tangent to curve

Pris parallel toPa.At airspeeds below V2andabove V1the aircraftcannot climb; the powerrequired for level flight ishigher than the poweravailable for climb.


68/96


Rate of climb

The equation for the rate of climb can bewritten as

Where a parabolic drag polar was substitutedfor CDand the rate of climb was assumedsmall.

The maximum climb rate is obtained whenVz/dV=0, which gives:

Vz =

Pa P

r

W=

1

WTV

1

2SV

3CD0

2KW

2

SV

#

$%&

'(

3SCD0

WV

4

2T

WV

2

4KW

S= 0


69/96


Solution for maximum climb

rate The quartic equation has four double roots. They

depend on the discriminant

The solutions are

AsDis always greater than 2T/W, the only positivesolution is

D =2T

W

!"#

$%&

2

43SC

D0

W

!

"#$

%&

4KW

S

!

"#$

%& = 2

T

W

!"#

$%&

2

+

3

Emax

2

V2=

2T

W D

!"#

$%&

/ 23SC

D0

W

!

"#$

%&

V2 = W3SC

D0

TW

+ TW

"#$

%&'

2

+

3

Emax

2

"

#$$

%

&''

where Emax

=

CL

CD

"

#$

%

&'max


70/96


Maximum climb rate

Substituting this value of airspeed in the climbrate equation we get

Where =EmaxT/W. Therefore, the maximum climb rate depends on:

Thrust available

Weight

Altitude

Wing surface

Vzmax

V=

1

3Emax

+ 2+ 3( ) 3

Emax

+ 2+ 3

( )


71/96


Climb rate requirements


72/96


Flight Dynamics 08

Greg Dimitriadis


73/96


Turning

The most general turn is a turnaccompanied by a change in height.

All the turn angles of the aircraft are

involved, roll, pitch and yaw In these lectures we will only cover

turns in a horizontal plane, i.e. nochange of height


74/96


Turn diagrams


75/96


Equations of motion

The equations of motion for the most

general case are:dx

dt= Vcoscos

dy

dt= Vcossin

dz

dt= Vsin

dV

dt=g

T

Wcos

D

Wsin

%&'

()*

d

dt=g

V

L

W+T

Wsin

%&'

()*cos cos%

&'()*

d

dt=

g

V

L

W+

T

Wsin

%&'

()*sin

cos


76/96


Constant speed horizontal

turn For a horizontal turn, =0.

For constant speed, dV/dt=0.

Also assume that there is no anglebetween the thrust line and theprojection of the flight path on theaircrafts plane of symmetry, i.e. =0.

The equations of motion become

Lcos =W, T =D


77/96


Load factor

The load factor is quite important for turns. Ifthe load factor is too great the pilot can beharmed or the aircrafts structure can bedamaged.

The load factor is defined as n=L/W.

From the equations of motion:

During a turn, the lift must balance not onlythe weight but also the centrifugal force. This

means that the loads acting on the aircraft arefar more important.

n = 1 / cos


78/96


Example

Assuming that the maximum load factor

allowed for an aircraft is nmax, calculate

the thrust required to perform a

horizontal turn of radiusRat a constant

airspeed V.


79/96


Solution (1)

We start with determining the load

factor. From the turning diagram, it is

seen that:

Now calculate the required lift. The lift

equation can be written as:

nW = mg( )2 + mV2

R!"#

$%&

2

nW =1

2SV

2CL


80/96


Solution (2)

The drag can be obtained, as usual, by

a drag polar, i.e.

The thrust is then given by

CD = CD0 + KCL2

= CD0 + K

2nW

SV2

"

#$%

&'

2

T = 12

SV2CD = 1

2SV2 C

D0

+ K 2nWSV

2"#$

%&'

2"

#$

%

&'


81/96


Solution (3)

This is the thrust required to perform the

required turn.

It must now be verified that the radius R

will correspond to a load factor lower

than nmax.

mg( )2 +mV

2

R

!

"#$

%&

2

W < nmax


82/96


The turning rate

For transport aircraft, the control ofturning rate to complete a turn in aprescribed time is important.

For military aircraft, a high turning rate

is vital (literally). The turning rate is defined as d/dt. The

last of the equations of motion gives (for=0, =0)

ddt

=gV

LW

sin


83/96


The turning rate (2)

Using the lift equation, the turning rate

can be calculated as

If we express the airspeed in terms of

the load factor and the lift coefficient we

get

d

dt

=

g

V

tan =g n 1

V

d

dt=g SCL

2Wn 1

n$%&

'()


84/96


Maximum turning rate

The equation for the turning rate contains thelift coefficient and the load factor

The lift coefficient can never be higher than

CLmax

.

The load factor can never be higher than nmax.

Therefore the maximum turning rate is

obtained by

ddt

"#$ %&'max

=g SCLmax

2Wnmax

1

nmax

"#$

%&'


85/96


The turning radius

The turning radius can also be

expressed as

Expressing the airspeed in terms of the

load factor and lift coefficient we get:

R =V

2

g tan=

V2

g n

21

R =2W

gSCL"#$

%&'

nn2 1


86/96


Thrust required for turning

The thrust equations can be written

directly as:

If it is assumed that the lift to drag ratio

is constant, then the thrust required for

turning is directly proportional to the

load factor.

Tr = nW

CD

CL


87/96


Power required for turning Multiplying the thrust equation by Vwe

obtained the power required for turning, i.e.

Using the drag polar to replace the dragcoefficient and the drag equation to write the

lift coefficient in terms of nand Vwe get:

Finally, replacing for nin this expression:

P =1

2SV

3C

D

P =1

2SV

3CD

0

+

2Kn2W

2

SV

P =1

2SV3CD0 +

2K

!

2

W

2

Vg2S

+

2KW

2

SV


88/96


Landing


89/96


Landing

The landing is performed in two phases:

Approach over a hypothetical obstacle to touch-

down

Ground run to a full stop Before reaching the obstacle the aircraft

makes an approach along the axis of the

runway with a glide angle between -2.5 and

-3.5 degrees. The approach speed is V2, i.e. 1.2Vstall.


90/96


Landing diagram

hrVTD V=0

V235ft

Approach Rotation (flare) Ground run Stop

V2

V2

Deceleration

x3= approach distance

x2= rotation distance

x1= deceleration distance

xg= ground run distance


91/96


Approach

The equations of motion are as for thedescent. For a generalized approach, sincethe angle is small:

whereE=CL/CD. The lift coefficient is fixed bythe second equation. The drag coefficient canbe obtained from the drag polar.

Therefore, for a chosen approach angle, the

first equation will yield the correct thrustsetting.

=1

E

T

W, and

1

2V

2

2CL =W


92/96


Approach (2)

If the height of the hypothetical object is

hoband the rotation height is hr, then the

approach distancex3is given by

And the approach time is

x3 =

hobh

r

tan

t3 = x

3

V2cos


93/96


Rotation

Rotation is a circular arc with radiusR. As in the take-off case, we have

Where nis the load factor.

The distance traveled during rotation is

And the time taken is

R =V2

2

g(n 1)

x2 = Rsin

t2 =V2

g(n 1)


94/96


Ground run

Once the aircraft has touched down theairspeed drops from VTDto 0.

The distance of the ground run can beapproximated by

Where is the mean deceleration:

xg =VTD2/2a

a

a =

0.30 0.35 for light aircraft with simple brakes

0.35 - 0.45 for turboprop aircraft without reverse propeller thrust

0.40 0.50 for jets with spoilers, anti - skid devices, speed brakes

0.50 - 0.60 as above, with nosewheel breaks

#

$

%%

&%

%


95/96


Parametric design

Design for performance is above all aprocess of optimization.

The aim is to satisfy or exceed allperformance requirements by finding theoptimal combination of parameters

The parameters are: Poweplant: Takeoff thrust, number of engines,

engine type of configuration

Wing: Wing area, Aspect Ratio, high-liftdevices


96/96

Flow diagram

conception aero performances

Documents