Class II Methodology for Drag Estimation

PRJ-22 Aerospace Design

Weight Estimation Using Class II Drag Calculation Method

Prof. Bento S. de Mattos

V20 – June 2009

Content

• Fundamental principles

• Zero-lift drag

• Induced drag

• Wave drag

• Class exercise

• Weight estimation

Total drag of the airplaneRef.: TORENBEEK, E. – “Synthesis of Subsonic Airplane Design”,

Kluwer, 1982 (pg. 368).

4

Total drag of the airplane

Prof. Mason, Virginia Tech


Prof. Mason, Virginia Tech

6


• Zero-lift, or parasite, drag

• Induced drag, or drag due to lift

• Drag due to compressibility, or wave

drag

• Total drag

CD,0 = CD,0,form + CD,0,friction

CD,i = kCL2 =CL

2/peA

CD,w= CD,w(M)

CD = CD,0 + CD,i + CD,w

7

Drag and Required Power

XDrag

Speed – V Speed - V

Induced Drag

Di ~ W2( b V)-2“Parasite” (viscous) Drag

Dp ~ Swet V2

Total Drag = Dp + Di

Power ( P)required =

D x V

V*prop V*jet V*prop

Power

V* = Optimum Speed to Fly for Maximum Range

Pavailable

Vmin Vmax

Power (P) = Thrust (T) x Speed (V)

8

CD0 CD0

CD0

CD0

CD0

CD0

CD0

9

Drag Breakdown of a Complete Airplane Configuration

10

Wing drag

Vertical tail drag

Flap drag

Nacelle drag Landing gear drag

Fuselage drag

Horizontal

tail drag

Aerodynamic drag components acting on

aircraft

11

Class I Method

Common Cfe Values

ref

wetfeD

S

SCC 0

12

Theoretical Background – Class II Method

A more useful measure of the parasite drag is the equivalent flat-plate drag

area, f . This quantity is exactly what it suggests–a flat plate of area, f , will

have the same drag as the airplane (when the plate is positioned

perpendicular to the wind). Thus, the total parasite drag is just

DP = f q

where q is the dynamic pressure.

You can find f by doing a component drag buildup. Each exterior component

of the airplane is considered separately, and the f of each is found. Then the

total f is determined by summing the component drag areas. In general, the

equivalent flat-plate area of the ith component can be computed from

ref

wet

iifref

iD

S

SQFC

Sf

C i

ii0

13

Theoretical Background – Class II Method

ref

wet

iifDS

SQFCC i

ii0

Friction coefficient

Form factor Interference factor

Area ratio

14

tw

,

2

,0 0.664 0.664

1 Re

2

f lam

x

xc

VxV

t

,

0,

0

1.328

Re

l

f lam

F lam l

l

c dx

C

dx

Local skin friction

coefficient

Average, or

integrated, skin

friction coefficient

Laminar flow over a smooth flat plate

X=0 X=l

Skin friction drag

15

laminar transitional turbulent

Laminar, transitional, & turbulent Flow

104 105 106 107 Rex

10

4

1

1000cfTurbulent

cf=0.74Rex-1/5

Laminar cf =1.328Rex-1/2

Transitional

X0

16

For 106 < Rel < 109 use:

Skin friction drag in turbulent flow

CF

106 107 108 109 Rel

0.0045

0.0035

0.0025

M=0

M=1

CF is small but the

dynamic pressure and

wetted area are large

65.0258.2

10 )144.01(Relog

455.0

MC f

17

DW,0 = DW,f + DW,p = zero-lift drag

DW,0 = CF,turbSwet q + k CF,turbSwetq

DW,0 = (1 + k) CF,turbSwet q

DW,0 = KWCFSwet q

Drag build-up by components

Consider, as an example, the drag build-up for the wing

(subscript w)

The drag estimate is based on a multiple of the friction

drag, KW, the form factor for the wing

18

Skin friction calculation Re, , /FC f M l k

Admissible surface roughness (Table 4.1.5.1-A):

l/k = reference length (in.)/surface roughness height (in.))

k(in.)=

0.02 to 0.08 x 10-3 polished metal

0.40 x 10-3 camouflage paint

6 x 10-3 dip-galvanized metal

106 107 108 Re,cut-off

106

105

104

103

l/k M=0 M=1

For Re<Re,cut-off

19

Skin friction drag in turbulent FlowCF

106 107 108 109 Rel

0.0045

0.0035

0.0025

M=0

M=1

Rel/l~2.4x106 per foot

at M=0.85 and h=35kft

Rel =Vl/ < Re,cut-off

CF

Reynolds Number

lV Re

Sutherland's formula can be used to derive the dynamic viscosity of an ideal gas

as a function of the temperature:

GasC [K] T0 [K]

μ0

[10−6 Pa s]

air 120 291.15 18.27

nitrogen 111 300.55 17.81

oxygen 127 292.25 20.18

carbon dioxide 240 293.15 14.8

carbon

monoxide118 288.15 17.2

hydrogen 72 293.85 8.76

ammonia 370 293.15 9.82

sulfur dioxide 416 293.65 12.54

helium 79.4 273 19

Valid for temperatures between 0 < T < 550 K with an

error due to pressure less than 10% below 3.45 Mpa.

Sutherland’s constant and reference

temperature for some gases.


Cut-off Reynolds Number

053.1)(21.38Re l/kcutoff Subsonic:

Transonic:16.1053.1)(62.44Re Ml/kcutoff

Source: Raymer

t

1

1/25.012 exp, rootWingwet ctSS

22

Wetted area Calculation

Wing

Taper ratioFrootref dchordSS exp

F

FF

d

l

23


Fuselage

2

3/2

,

11

21

FF

FFFuswet ldS

p

24


Engine Nacelles

25


Engine Nacelles

ppplugwet DlS p7.0,

3/5

, 18.0113

11

g

g

g

eg

gggengaswetl

D

D

DDlS p

n

ef

n

l

nn

hll

n

lnncowlingfanwet

D

D

l

l

dl

Dl

l

lDlS 115.18.035.02,,

Wetted Area


Component Approximate Wetted Area Aw

Form Factor

28

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

k

Fineness Ratio λ=Length/diam.

kfactor in fuselage drag

correlation

NAA

Datcom

Torenbeek

Hoerner

Fuselage form factor as determined by

different investigators

(Kf=1+k)

Form factor

Fuselage:

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930090976_1993090976.pdf

New link for NACA Technical Report 824.

4003

601

f

ffus

F

Body with blunt base

Closed body (CD,b=0)

d

base

fus

fusfus D

ref

wet

fusfusfD CS

SQFCC

0


30

Wing form factor as determined by

different investigators

k - factor in wing drag correction

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.05 0.1 0.15 0.2

Thickness Ratio t/c

k

Jenkinson

Torenbeek

NAA

Hoerner

DATCOM 2

DATCOM 1

(Kw=1+k)

cos34.1/100/6.0

128.018.04

m

t

W Mctctx

FF

Form factor

Wing:

Chordwise location of the maximum thickness (should be used a actual length in m)

Nacelle:

Sweep angle of the wing measured at the line that is generated by joining the maximum thickness location of the airfoils (xt)


New link for NACA Technical Report 824.


Interference Factor

Interference Factor

Interference Factor


Interference Factor


Roskam suggests that the value for the wing-fuselage interference factor should be

taken from the graph below

Interference Factor

Interference Factor

Estimated values of interference drag originating in the corners of various

tail configurations

Source: Horner, Fluid-dynamic Drag

Miscellaneous Parasite Drag

The drag of some items does not quite fit the form

must be computed in an alternative manner. Table below gives values of f /Afrontal,

the equivalent flat plate drag area normalized by the projected frontal area, of

landing gear components. Deployed flaps have a drag area given roughly by

with bflap representing the span of the flap and δflap the flap angle in degrees.

Recall that takeoff flap setting is approximately 25◦ whereas a δflap of 50◦ is

used on landing. Fuselage-mounted speed brakes have f /Afrontal = 1, while

wing-mounted speed brakes, or spoilers, have f /Afrontal = 1.6.

Estimated equivalent flat-plate drag

areas for landing gear components.

(From Raymer.)


Drag Caused by Flap Deflection


Definition of flap drag @ 0.70 CL max

41

DCD,flaps = 0.9 (cflap /c)1.38 (Sflap /S) sin2 d (slotted flaps)

Alternative Formulation for the Calculation of

Drag Coefficients for Flaps




Windmilling Engine and Propeller Drag

When computing engine-out performance, it is necessary to include additional

drag from stopped or windmilling engines or propellers. While operating, these

propulsion components are considered to have no drag since the reported thrust

includes a decrement for forces generated in the drag direction. Propeller drag

depends on the solidity, σ, given by

where B is the number of blades, cavg is the average blade chord, and R is the

propeller radius. Note that this quantity just equals the ratio of blade area to

disk area. The drag area is given by

A windmilling jet has an equivalent drag area given by 0.3 times the face area.


Landing gear Drag


(SI)

nits)(British U

785.0

785.0

S

KM

qS

D

S

KW

qS

D

Analysis of flight measured drag, for a number of civil transport aircraft, indicates that the

following values of K should be used

In British units with W in lbf

deflection flap fullfor 108.1K

deflection flap zerofor 103.3

3

3

x

xK

In SI units with M in kg

deflection flap fullfor 1031.0K

deflection flap zerofor 1057.0

3

3

x

xK

The reduction from K with flap deflection is

assumed to be linear

nits)(British U 105.13.3 3

x

MaxF

FKd

d

Source: ESDU

For the calculation of landing gear dreag when the geometry is unknown a empyrical formula

can be used

45

CDi CDi

CDi

CDi

CDi

CDi

CDi

46

What is?

•Induced drag is the drag related to the generation of lift.

•It results from the angle of attack induced from the 3-D flowfield.

•For an elliptical wing:

•For a non-elliptical wing, we have introduced Oswald's Efficiency Factor, e

•So is a measure of the reduction in efficiency over the optimum elliptical wing

case.

•In other words, the elliptical lift distribution is optimum in terms of induced

drag.

A

CC L

Dip

2

Ae

CC L

Dip

2

47

Oswald’s Factor e

M = Mach number

= Taper ratio of the reference wing

ne = number of engines placed under the wing

A = Aspect ratio

(t/c) – averaged maximum thickness ratio of the wing

25 = Sweepack angle at ¼ chord


48

Induced drag @ supersonic regime


49

Reducing Induced Drag


Winglets

Wings of higher aspect ratio

50



Ae

CC L

Dip

2

d

1

1e

d

Source: Anderson, Aircraft Performance and Design.

Swept wings

Induced Drag

012

2

CCCCC

C LL

Di

p

d

1

11

2Ce

Source: Schlichting. H. und Truckenbrodt , Aeodynamik des Flugzeuges, Zweiter Band.

Induced drag of symmeric twisted trapezoidal wing

a) Planform with linear twist

b) Induced drag of the untwisted wing

c) and d) Twist contribution to induced drag

a

b

c

Aspect ratio

d

1

1

C

2

1

0

C

2C

angle induced1

52



Aircraft L/Dmax Wing Span (m) Aspect Ratio

WWII Bomber

Boeing B-29 16.8 42.98 16.8

B-24 J Liberator 12.9 33 11.5

Boeing B-17G 12.7 31.64 7.58

Martin B-26F 12 21.64 7.66

WWII Fighter

Lockheed P-38L 13.5 15.84 8.26

P-51D Mustang 14.6 11.28 5.86

Me 262A 14.09 12.53 7.23

53

CDwave CDwave

CDwave

CDwave

CDwave

CDwave

CDwave

54

Wave drag @ supersonic regime

Example: CDWave Comparison With Two Methods for a SSBJ

• Fuselage length = 198 ft

• Height of Airplane with landing gears down = 33 ft

• Span of the delta wing = 103.17 ft

• Wing aspect ratio

= 2.6 with wingtips Up

= 1.9 with wingtips Down

55

Rallabhandi and Mavris’s Approach

Below are given an analytical expression

for the wave drag assuming the aircraft

body to be Sears-Haack body.

Wave drag @ supersonic regime

wet

DSl

VolC

4

2

wave

128

p

British System

56

Induced drag @ supersonic regime

Raymer’s Approach

• Assumptions:

-Correlates aircraft wave drag to an

equivalent Sears-Haack body at Mach = 1.2.

57

Wave drag @ supersonic regimeExample:

CDWave calculation for a SSBJ fuselage by the two methods

from previous slides

ft3

ft

58

Exercise

ERJ 145 CD0 Calculation for the Wing and Fuselage

• Wing reference area = 51.12 m2

• Wing taper ratio = 0.2543

• CMA of the wing = 2.9 m

• MMO @ 37000 ft = 0.78

• Fuselage length = 27.93 m

• Fuselage diameter = 2.28 m

• Sweepback angle @ 1/4 chord = 22.72º

• Chord @ root = 4.09 m

• (t/c)root = 14%

• (t/c)tip = 9.5%

• (t/c)averaged = 12%

General Data


Calculation will be performed for the cruise

condition (Number of Mach = 0.78)!

60

ERJ 145 CD0 Calculation for the Wing

Friction coefficient is given by

We consider in this example that the laminar portion of the wing covers

10% of its whole exposed area (usually this figure is around 5%). That

means we must take klam = 0,10.

Calculation of the Reynolds number

7

6

1025.11027.18

9.2230343.0Re x

mCMAV

smkg

sm

cruise

turbflamlamflamf CkCkC ,, )1(

716.1

053.1

16.1053.1 107.678.0003.0

290062.44)(62.44Re x

mm

mmMl/kcutoff

We take 1.25x107 because is the lowest Reynolds number!


61


Friction coefficient is given by

We obtain the Cf,lam and the Cf,turn

The friction coefficient is the given by

turbflamlamflamf CkCkC ,, )1(

333

,, 1051.21075.29.010345.01.0)1( xxxCkCkC turbflamlamflamf

3x

7, 10 0.3756

10 x 1.25

328.1

Re

328.1 lamfC

3

65.0258.27

10

65.0258.2

10

, 1075.278.0144.011025.1log

455.0

144.01Relog

455.0

xM

C turbf



We consider the interference factor QW=1.01 (from the Roskam’s Graph)

The form factor of the wing can be calculated using the following

expression

With Xt = 0,35xCMA = 0,35 x 2,90 m = 1.1015 m

The wetted area is given by

The area ratio can easily be calculated:

33

,0 1096.501.17.1384.110 x 51.2 xC wingD

7.112.51

8.862

2,

m

m

S

S

ref

wwet

384.15.15cos78.034.112.010012.0015.1

6.01

28.018.04

o

WFF

28.018.04cos34.1/100/

6.01 m

t

W Mctctx

FF

2

exp, 8.862543.01

2543.04736.1114.025.018.412

1

1/25.012 mctSS rootWingwet

t


ERJ 145 CD0 Calculation for the Fuselage

The Reynolds number for the fuselage is

Re is lower that the cut-off Reynolds number. Thus, the usual Reynolds number

must be taken into the equations for the calculation of Cf,lam and Cf,turb.

Considering that the fuselage presents a rough surface, the cut-off Reynolds number must also

be calculated to taken this characteristic into account

(Raymer)

8

6

310206.1

1027.18

93.27230343.0Re x

lV

mskg

msm

m

kg

fuscruise

816.1

053.1

16.1053.1 1031.378.000635.0

2793062.44)(62.44Re x

mm

mmMl/kcutoff

3

65.0258.28

65.0258.2,

10962.178.0144.0110206.1log

455.0 ...

144.01Relog

455.0

xx

MC turbf

4x

8, 10 1.209

10 x 1.206

328.1

Re

328.1 lamfC

334

,, 10778.110962.19.010209.11.0)1( xxxCkCkC turbflamlamflamf

ERJ 145 CD0 Calculation for the Fuselage

The form factor of the fuselage can be calculated by

The wetted area can be obtained by using a Torenbeek’s formulation

(DATCOM 78)

We consider the interference factor QF=1.

The area ratio can easily be calculated: 523.312.51

8.862

2,

m

m

S

S

ref

fwet

33

,0 1076.6 523.308.1110778.1 xxC fuselageD


65

Preliminary Weight Estimation


66

2. Preliminary Weight Estimate

WTO = WE + WTFO + WPLC + WF,USED + WF,RES

=Take-off Weight

WE =Empty Weight

WF = WF,USED + WF,RES

= Weight of Fuel Used+ Weight of Fuel Reserve

= Total Fuel Weight

WPLC =WPL+WCREW = Weight of Payload +Weight of Crew

MTFO = WTFO / WTO=(Trapped Fuel and Oil Weight)/WTO

MFUEL = WF/WTO= Fuel Fraction

Commercial Airplane Design 67

Solve for the empty weight knowing WPLC

WE = (1 – MTFO – MFUEL)WTO – WPLC = aWTO + b

WTO

WE

0

-WPLC

(1-MTFO-MFUEL)

increasing

Empty weight vs take-off weight relation

Fuel fraction

needed for

mission,

including

reserves

68

WF = WTO – WFINAL=WTO – (Weight at End of Mission)

WF/WTO = MFUEL= 1 – WFINAL/WTO = 1 – MFINAL

Fuel Needed for Mission

1 2 3

4

5 6

78

9

10

11

Mission Profile

Normal

Diversion


69

1 2 3

4

5 6

78

9

10

11

Mission Profile

0.99 0.99 0.995

exp[-RCj/V(L/D)] exp[-Cj/(L/D)] exp[-Ralt.Cj/V(L/D)]

0.98 0.99 0.98 0.99

0.992

Segment weight fractions Wi / Wi -1

Cruise

Loiter


70

11

10 1

nFINAL i

FINAL

iTO i

W WWM

W W W

MFINAL =(W11/W10)(W10/W9)(W9/W8)….(W2/W1)(W1/W0)

,

, ,1F USED

F USED FINAL F RES

TO

WM M M

W

,

,

LAND NOM

FINAL F RES

TO

WM M

W

5 9,

,

1 61 1

1F RES i i

F RES

i iTO i i

W W WM

W W W

Final Weight Fraction

Fuel Weight

Fraction Used

Nominal Landing Weight

Reserve

Fuel

Fraction

71

L/D Calculation

Cruise!


72

L/D Calculation

For a jet airplane the L/D for maximum endurance (to applied for the loiter phase) can be simply obtained by


73

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 2000 4000 6000 8000 10000

Range (mi)

1-M

fin

al

Normal Mission Fuel Fraction vs Range

This is the nominal value of the ratio WF,USED/WTO

1-MFINAL = 0.00316(R-800)1/2

74

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 2000 4000 6000 8000 10000

Range (miles)

1 -

Mfi

na

l+M

res

Total Fuel Fraction vs Range

1-MFINAL+MRES=0.0048R1/2

Nominal ratio of total fuel carried to take-off weight, MFUEL

75

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0 200,00

0

400,00

0

600,00

0

800,00

0

1,000,0

00

1,200,0

00

1,400,0

00

Take-Off Weight (lbs)

Mtf

o

Weight Fraction of Trapped Fuel & Oil

MTFO=0.227(MFUEL)2/3(WTO)-1/3

Correlation for the weight fraction of trapped fuel and oil

76

Correlation of empty weight vs take-off

weight for 45 airliners

0.0

100.0

200.0

300.0

400.0

500.0

600.0

700.0

0.0 200.0 400.0 600.0 800.0 1000.0 1200.0 1400.0

Take-off weight, Wto (klbs)

Em

pty

we

igh

t, W

e (

klb

s)

Actual weights

logWe=(logWto - A)/B

We=0.5Wto


77

Correlation of empty weight vs. take-off

weight for 45 airliners

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.0 200.0 400.0 600.0 800.0 1000.0 1200.0 1400.0

Take-off weight, Wto (klbs)

Em

pty

we

igh

t fr

ac

tio

n, W

e/W

to

We/Wto = 1.59/(Wto/1000) .̂0906


78

WE=aWTO-WPLC

Historical

correlation

WE=0.504WTO

WE

0

-WPLCWTO

Market survey aircraft

Estimating Aircraft Empty Weight

Estimating Cruise Fuel Consumption

Performance

Max operating Mach number 0.83

Max operating altitude 41,000 ft (cabin altitude: 8,000 ft)

Take-off field lenght 6,500 ft (SL / ISA + 15°C / MTOW)

Landing field 5,000 ft (SL / MLW = 90% of MTOW)

Range with max payload 2,200 nm (overall fuel volume for 3,200 nm version)

External noise FAR 36 Stage IV minus 15 db

IPET7 Airliner

80

Estimating Cruise Fuel Consumption

41000 ft

0,150

0,170

0,190

0,210

0,230

0,250

0,270

0,290

0,40 0,50 0,60 0,70 0,80 0,90

Mach

SR

[n

m/k

g]

MTOW 90% MTOW 80% MTOW

Long Range MMO

SR vs. Mach number 41000 ft

0,00

2,00

4,00

6,00

8,00

10,00

12,00

14,00

0,40 0,50 0,60 0,70 0,80 0,90

Mach

M*L

/D

MTOW 90% MTOW 80% MTOW

Mach*L/D vs. Mach number

The number of Mach for maximum specific range (SR) is not the same as that for

maximum M*L/D because sfc increases with speed

IPET7

IPET7

Class II Methodology for Drag Estimation

Documents

total drag

total parasite drag

induced drag wave

total f

f qwhere q

equivalent flatplate

suggestsa flat plate

laminar cf