3.3Aircraft Drag

8/14/2019 3.3Aircraft Drag

1/36

Aircraft Drag IIDEN/305

G. Dorrington Oct. 2001


2/36

Drag Coefficient

D = 1/2r U2 S CD

= q S CD

where D = Drag

q = dynamic pressure = 1/2r U2

U = velocity

r = density

S = wing planform area


2s = b

c


3/36

Drag Coefficient of an Aircraft

The total drag coefficient of an aircraft (with lift)

can also be written as:

CD = CD0 + kPCL2 + CL2/(e p A)

or

CD = CD0 + k CL2/(p A) where k is typically about1.2



4/36

Drag of Aircraft

The Drag of an aircraft in cruise where

L=W can hence be written as:

D = q S {CDo + k CL2/(p A)}

= q S CDo + q S k (L/qS)2 /(p A)

= q S CDo + k W2/(qS p A)= q S CDo + k W

2/(q b2p)



5/36

Drag Polar


`

U

D

D = (rU2/2) S CDo + 2kW2/(rU2b2p) = AU2 + B/U2

induced

profile

total


6/36

Zero-lift (or Profile)

Drag of Aircraft

CD0 = KMIS CD0i Si /S ref

mutual interference = 1.3


7/36


Drag of Aircraft

Dprofile = qSCD0 = KMFS qCD0i Si

S = Sref= planform area of wing


8/36


Drag of Aircraft

Dprofile/(qS) = CD0 = KMFS CD0i Si/S

S = Sref= planform area of wing


9/36

Need for wind tunnel tests:

empirical resultsComputational methods cannot yet predict profile drag of

any body above Re=500,000

Need to do wind tunnel testing: Re scaling

Drag estimation is still a black art

Following are some general empirical results


10/36

Fuselage: Bodies of Revolution

Which body has the

lowest drag based on

frontal area pd2/4

(same for all) ?

d

L


11/36


12/36


L/d around 3-4 is best

d

L


13/36



14/36



15/36


Based on wetted area, empirical equation:

CD0 = Cf(1 + 1.5 (d/L)1. 5 + 7 (d/L)3 )

where wetted area is approximately 0.85 p d l

L

d


16/36


Based on wetted area, empirical equation:

CD0 = Cf(1 + 1.5 (d/L)1. 5 + 7 (d/L)3 ) = Cf x form factor

where wetted area is approximately 0.85 p d l

L

d

L/d form factor

2 2.41

3 1.55

4 1.30

5 1.19


17/36


For turbulent flow

Cf= 0.455 (log 10 ReL )-2.58

ReL = r U L/ m

For example, at ReL = 8000,000 , Cf= 0.0031

Re. No. Cf turbulent

1000000 0.0045

2000000 0.0039

4000000 0.0035

8000000 0.0031

16000000 0.0028


18/36


Hence if L/d = 3 :

CD0 = 0.0031 x 1.55 = 0.0048

based on S = 0.85 p d l

L

d


19/36


Hence if L/d = 3 :

CD0 = 0.0031 x 1.55 = 0.0048 based on S = 0.85 p d l

3m

1m


20/36


Hence if L= 3m, d =1m:

CD0 = 0.0031 x 1.55 = 0.0048 based on S = 8 m2

Dprofile /q = 0.0048 x 8 m2 = 0.038 m2 (about 20 cm x 20 cm)

3m

1m


21/36


Hence if L= 3m, d =1m:

CD0 = 0.0031 x 1.55 = 0.0048 based on S = 8 m2

Dprofile /q = 0.0048 x 8 m2 = 0.038 m2 (about 20 cm x 20 cm)

3m

1m


22/36


Hence if U = 39 m/s , at sea level:

Dprofile = 36 N , Pprofile = Dprofile U = 1.4 kW

3m

1m


23/36

Wing Profile Drag


24/36

Wing Profile Drag

t

c

CD0 = Cf(1+ 2 t/c + 60 (t/c)4)

based on wetted area = 2.2 S

hence if t/c =0.15, CD0 = 1.33 Cf


25/36

Wing Profile Drag

if t/c =0.15, CD0 = 1.33 Cf

and if the flow is fully turbulent

Cf = 0.455 (log10 Rec)-2.58

where Rec = r U c / m

t

c


26/36

Wing Profile Drag

if t/c =0.15, CD0 = 1.33 Cf

and Reynolds number = 1000,000

CD0 = 0.0045 x1.33 = 0.006

based on wetted area Si = Swet = 2.1 S

so based on S = Splanform

CD0 = 0.006 x 2.1 = 0.0125


27/36

Wing Profile Drag

if t/c =0.15, CD0 = 1.33 Cf

But flow is fully laminar

Cf = 1.328 Rec-1/2

where Rec = r U c / m < 300,000

t

c


28/36

Wing Profile Drag

if t/c =0.15, CD0 = 1.33 Cf

and Rec=300,000fully laminar

Cf = 1.328 Rec-1/2 = 0.0024

CD0 = 0.0032 based on wetted area

= 0.0068 based on planform area

t

c


29/36

Wing Profile Drag

if t/c =0.15, CD0 = 1.33 Cf

but flow is mixed laminar and turbulent

Laminar BLTurbulent BL

Transition


30/36

Wing Profile Drag

Laminar BLTurbulent BL

Transition

Cf S = 0.455 (log10 Rec)-2.58 S

- 0.455 (log10 Rex)-2.58 S x/c

+ 1.328 Rex -1/2 S x/c

x Rex = Recrit = 300000


31/36

Wing Profile Drag

At Rec= 600,000

Cf S = 0.0049 S

- 0.0057S x/c

+ 0.0024 S x/c


x = Recritm/rU

c = Recm/rU

x/c= Recrit /Rec

= 0.5


32/36

Wing Profile Drag

At Rec = 600,000

Cf S = 0.0031 S

- 0.0057S /2

+ 0.0024 S /2



33/36

Wing Profile Drag

At Rec = 600,000

Cf S = (0.0031 - 0.00165) S = 0.00145 S



34/36

Total Profile Drag

D = q S CD0 = KMI {q S C D0 wing+ q S tail C D0 tail

+ q S fuselage C D0 fuselage ..}


35/36

Design ImplicationsIf significant natural laminar flow is feasible, then

note advantage to keep chord c low to maximise

laminar chord fraction x/c

Implies high Aspect Ratio A = b2

/S

To reduce profile drag:

reduce wing area S

reduce wetted area of fuselage and control surfacesetc., keep streamline - reduce parasitic drag (rivets

etc.), keep surfaces smooth

keep aircraft small


36/36

End

See exercises in gstutor

G Dorrington Oct 2001

3.3Aircraft Drag

Documents

Transcript of 3.3Aircraft Drag