For most objects moving througha fluid, the significant fluid force is drag.

However for some specially shapedobjects the lift force is also important.

LIFT

LIFT

DRAG

TIME OUT

Newtonian Theory (1687) entire second book of Principia dedicated to fluid mechanics - assumed particles of fluid lose momentum

normal to plate but keep momentum parallel to plate.

p due to random motion of molecules

p p

Aside

Force normal to plate, N = Time rate of change of the normal component of momentum =(mass flow) x change in normal component of velocity =

N = ( V A sin) x (V sin)N/A = p – p = (V sin)2

(p - p ) / (1/2 V 2) = Cp = 2 sin2

Cp = 1.86 as M Aside

Forces on airplane atlevel speed and constantheight and speed.

Lift force is the component of R that is perpendicular tofree stream velocity, and drag is the component of R parallel to the free stream velocity. If planes height is not changing then: Lift = Weight

Force generated if we brought fluid directly approaching area to rest

CL = FL/(1/2 V2Ap)

CD = FD /(1/2 V2A)

FL

FD

Note: FL is not parallel to N

Chord of an airfoil is the straight line joining the leading and trailing edge. If mean line and chord do not coincide thenairfoil is cambered. Ap refers to planform area, maximum projectionof wing (independent of angle of attack, )

Ap and c are independent of

CL =FL/(1/2 V

2Ap)

CD =FD/(1/2 V

2A)

Ap = planform areamax. proj. of wing

A

Given: Kite in standard air, mass = 0.2 kg;CL = 2sin(); CL/CD = 4. Find

U= 10 m/s 0.2kg (g)

Area = 1 m2

5o

= ?

Kite in standard air, CL = 2sin(); CL/CD = 4

FL

UFD

T

mg

= ?

CL = FL/(1/2 V2Ap)

CD = FD /(1/2 V2A)

CL/CD = FL/FD = 4(Ap = A)

9.144

Fy = FL – mg –Tsin() = 0Fx = FD –Tcos() = 0FL = CL A ½ U

2

CL = 2sin(5o) = 0.548FL = 33.7NFD = FL/4 = 8.43N

FL

UFD

T

mg

= ?

tan () = Tsin()/Tcos() = (FL – mg)/FD

= tan-1{{(FL – mg)/FD} = 75.1o

FL

UFD

T

mg

= ?

For all cases angle of attack is 4o and aspect ratio is 6.Lift to drag ratios of about 20 are common for modern transport planes.

flat plate bent plate airfoil

IMPORTANCE OF CAMBER

Otto Lilienthal on a monoplaneglider in 1893

Otto Lilienthal on a biplaneglider in 1893

Otto Lilienthal (1848-96) is universally recognized as as the first flying human. His wings were curved (camber – height = 1/12th of chord). On August 9th, 1896 Lilienthal suffered a fatal spinal injury, falling10-15 meters from the sky.

Lift = U

2-D potential flowLper unit span = U

= C vds

“In order for lift to be generated there must be a net circulation around the profile.”

Pg 448 ~ our book

ASIDE

INVISCID INCOMPRESSIBLE FLOW AROUND AN 2-D CYLINDER

ASIDE

p + (/2)U2 = p + (/2)(2U sin)2

(2U sin) comes from 2-D potential flowfor a source, sink and free stream flows

FREE STREAM ON SURFACE

ASIDE

Governing equation for incompressible, irrotational flow:

•V = 0; x V = 0; V = ; 2 = 0

Laplace’s Equation, 2 = 0, is a second order, linear partial differential equation.

Solutions can be added!

Turns out that a 2-D source, 2-D sink and free stream makes a flowlike that over a 2-D ellipse.

ASIDE

m/(Ua) =1

Potential flow solutions: Source + Sink + Uniform flow

ASIDE

Doublet: 2(a)m =

constant as a shrinks to zero

ASIDE

add circulation: u = ro

p + (/2)U2 = p + (/2)(2U sin + ro)2

= C(v+u)ds = 2rou

ASIDE

Lift is integral of pressure around cylinder

= C vds = ro2ro; Lper unit span = U

ASIDE

Bernoulli’s Equation

We have just shown thatif there is a net circulation,

then there is lift.

SO WHAT IS WRONG WITH THIS PICTURETAKEN FROM A POPULAR TEXTBOOK?

Both figures claim lift, which figure’s streamlines are consistent with lift?

U = 4 m/sR = 7.7 cmRe = 4 x 104

= 0

U = 4 m/sR = 7.7 cmRe = 4 x 104

= 4U/R

Both develop lift, see streamlines pinched on top (faster speeds, lower pressure)and wider on bottom (lower speeds andhigher pressure)

Kelvin’s theorem showed that the circulation around any closed curve in an inviscid, isentropic fluid is zero. Consequently there must be circulation around the airfoil in which the magnitude is the same as and whose rotation is opposite to that of the starting vortex.

A CONSEQUENCE OF CIRCULATION AROUND WING IS STARTING VORTEX

U = 30 cm/sChord = 180 mmRe = 5 x 105

Floating tracer method

Starting vortex

“Trailing vortices can be very strongand persistent, possibly being a hazard to other aircraft for 5 to 10 miles behinda large plane – air speeds of greater than 200 miles have been measured.”

Bernoulli’s Equation

HINT: SAME THING THAT IS WRONG WITH THIS PICTURE?

“The phenomenon of aerodynamic list is commonly explained by the velocity increase causing pressure to decrease (Bernoulli effect) overthe top surface of the airfoil..” ~ YOUR BOOK PG 448

Norman Smith: Physics Teacher, Nov. 1972, pg 451-455.

“In spite of popular support, Bernoulli’s Theorem is not responsible for the lift on an airplane wing.”

“The phenomenon of aerodynamic list is commonly explained by the velocity increase causing pressure to decrease (Bernoulli effect) overthe top surface of the airfoil..” ~ YOUR BOOK PG 448

Lift is a result of Newton’s 3rd law. Lift must accompany a deflection of air downward.

BERNOULLI EQUATION GOOD FOR STREAM TUBES WHEREENERGY IS NOT BEING ADDED OR SUBTRACTED

Yet one can argue that B.E. is valid

for outer stream tubesso book not wrong.

BERNOULLI’S EQUATIONX-MOMENTUM EQUATION:INVISCID and NO BODY FORCES:

(Du[t,x,y,z]/dt) = - p/x (u/t) + u(u/x) + v(u/y) + w(u/z) = - p/x

STEADY: u(u/x) + v(u/y) + w(u/z) = - p/x

dx[u(u/x) + v(u/y) + w(u/z) = - p/x]

aside

BERNOULLI’S EQUATION

CONSIDER FLOW ALONG A STREAMLINE:

ds x V = 0

udz-wdx = 0; vdx-udy = 0

u(u/x)dx + v(u/y)dx + w(u/z)dx = - p/xdx

u(u/x)dx + u(u/y)dy + u(u/z)dz = - p/x dx

aside

BERNOULLI’S EQUATION

u(u/x)dx + u(u/y)dy + u(u/z)dz = - p/x dx

u{(u/x)dx + (u/y)dy + (u/z)dz} = - (1/)p/x dx

du

udu - (1/) p/x dx ½ d(u2) = - (1/) p/x dx

aside

BERNOULLI’S EQUATION

X-MOMENTUM EQUATION: ½ d(u2) = - (1/) p/x dxY-MOMENTUM EQUATION: ½ d(v2) = - (1/) p/y dyX-MOMENTUM EQUATION: ½ d(w2) = - (1/) p/z dz

½ d(V2) = - (1/) dpdp = - V dV Euler equation

u2 + v2 + w2 = V2

p/x dx + p/y dy + p/z dz = dp

aside

BERNOULLI’S EQUATION

½ d(V2) = - (1/) dp {½ d(V2) = - dp}

INCOMPRESSIBLE:

½ (V22) - ½ (V1

2) = - (p2 – p1)p2 + ½ (V2

2) = p1 + ½ (V12)

= constant along streamline

If irrotational each streamline has same constant.

aside

BERNOULLI’S EQUATION

p2 + ½ (V22) = p1 + ½ (V1

2)

Momentum equation and steady, no body forces, inviscid, incompressible along a streamline.

Kinetic Energy / unit volume

If multiply by volume have balance between work done by pressure forces and change in kinetic energy.

(interesting that for an incompressible, inviscid flow energy equation is redundant for the momentum equation)

aside

LIFT ‘Measurements’

Lift = U= unsymmetrical flow field

p2 + ½ (V22) = p1 + ½ (V1

2) = constant along streamline

CL = FL/(½ U2Ap)= f(shape, Re, Ma, Fr, /l)

Except at very low Rew does not contribute directly to lift

Calculated (dots) and measured (circles) pressure coefficients for airfoil at = 7o.

= (p-p)/(1/2 U2)

Juan Lopez, which curve is for the upper surface?

Calculated (dots) and measured (circles) pressure coefficients for airfoil at = 7o.

= (p-p)/(1/2 U2)

_________ pressure gradient

__________ pressure gradient

Calculated (dots) and measured (circles) pressure coefficients for airfoil at = 7o.

= (p-p)/(1/2 U2) unfavorable pressure gradient

favorable pressure gradient

First person in 3rd row, which surface iis flow likely to separate?

Stall results from flow separation

over a major section of the upper surface

of airfoil

Rec = 9 x 106

CL = FL/( ½ V2Ap)

Because of the asymmetry of a

cambered airfoil, the pressure

distribution on the upper and lower surfaces are different.Must have camber to

get lift at zero angle of attack.

Rec = 9 x 106

CL = FL/( ½ V2Ap)

Rec = 9 x 106

CL = FL/( ½ V2Ap)

Typical lift force is of the order unity. (dynamic pressure x planform area)Consequently, FL ~ ½ V2Ap

Wing loading = FL/Ap

= 1.5 lb/ft2 1903 Wright Flyer= 150 lb/ft2 Boeing 747= 1 lb/ft2 bumble bee

As angle of attack increasesstagnation point moves

downstream along bottom surface, causing an

unfavorable pressure gradient at the nose*.

*

*

stagnation point

Stream tube narrows then widens

unfavorable

favorable

favorable to unfavorablecauses lam. to turb. trans.

= 2o

*

= 10o

*

*

= 15o -

= 15o +

separation at leading edge

Check angle =15*

Transition from laminar to turbulent flow on

upper surface.

Laminar flow sections designed to fly at low

angles of attack,but has less

maximum lift.Re = 9 x 106

Turbulent Lam.Turbulent

CD = FD/( ½ V2Ap)

CL = FL/( ½ V2Ap)

Lift-Drag Polars are often used to present airfoil data.

Plot is for one particular Rec number

X

X is where you find the most efficient angle of attackHigh performance airfoils can generate lift that is 100 times Their drag – so can flide 100m for every m they drop in altitude.

CL related to load

CD related (mustinclude fuselage drag,

etc.) to drag thatplane must overcome

to achieve lift.

Higher the CL/CD the better!

LIFT – DRAG POLAR

Graph for one Re #different angles

of attack

Different angle of attack

Note that x and y axisHave different scales

Wing tip vortices

crop duster

All real airfoils of finite span, wings, have more drag and less lift than what 2-D airfoil section would indicate.

Trailing vortices reduce lift because pressure difference is reduced.

LIFT = U

U

Two primary leading edgevortices made visible by

air bubbles in water.(Van Dyke Album of Fluid Motion)

Schematic of subsonic flow over the top of a delta wing at an angle

of attack.

The tendency for flow to leak around the wing tipsgenerally cause streamlines over the top surface ofthe wing to veer to the wing root and streamlinesover the bottom surface veer to the wing tips.

Endplates (winglets) at end of wing reduces tip vortex

Winglet is a vertical or angled extension of the wing tips for reducing lift-induced drag.

CD = CL CL2/(ar)

ar = b2/Ap

Winglets work by increasing the effective ar of the wing without increasing the span.

The vortex which rotates aroundfrom below the wing strikes thewinglet, generating a small lift force.

The tendency for flow to leak around the wing tipsalso produces wing tip vortices downstream of the wing which induce a small downward component of air velocity in the neighborhood of the wing itself.

Not all same strength

Lift force is perpendicular to local relative wind.eff is angle that wing sees, angle between chord line and relative wind.

Local lift vector is aligned perpendicular

to the local relative wind.

“This causes the lift force to lean backwards a little, resulting in some of the lift appearing as drag.” Fox et al.

V=V

Di = L sin(i) ~ L i (or L )i ~ CL/( ar) [theory/exp]

CD,I ~ CL i ~ CL2/( ar)

* From Fundamentals of Aerodynamics by Anderson

*

• Induced drag was derived from inviscid, incompressible flow theory –

• Induced drag only for finite wing

• No skin friction or separation

• D’Alembert’s paradox does not occur for finite wing!

Loss of lift and increase in drag caused by finite-span effects are concentrated near the tip of the wing; hence short stubby wingswill experience these effects more severely than a very long wing.

“New” glider by Wright brothers which was astoundingly successfulhad an increase in wingspan to chord ratio from 3 to 6.

Expect induced drag effects to scale with wing aspect ratio = b2/Ap

ar = b/cc

ar = b2/Ap

L/D ~ 400 for ar = L/D ~ 40 for sailplane with ar = 40L/D ~ 20 for typical light plane

with ar = 12

Soaring birds(but large wings have more inertia,

harder to turn quickly)

Maneuvering birds

Tuna Butterfly Fish

Pike

Bass

Di = L sin(i) ~ L i (or L ) ~ CL/( ar)

( = 0; ar = )Di = D = L = LCL/( ar)

CD = CL2/( ar)

CD = CD, + CD,i = CD, + CL2/( ar)

FINITEWING

CD = CD, + CD,i = CD, + CL2/( ar)

CL = FL/( ½ U2Ap) and W = or ~ L

FOR AIRCRAFTCD = CD,0 + CD,i = CD,0 + CL

2/( ar)CL = FL/( ½ U2Ap) and W = or ~ L

FOR WING

FOR INFINITE WINGCD, = FD/( ½ U2Ap)CL, = FL /( ½ U2Ap)

FROM DATA FIGURES

To get same lift (same CL) as infinite armust increase ~ CL/( ar); [linear]

To get same drag (same CD) as infinite ar must increase CD ~ CL = CL

2/(ar); [quadratic]

EXAMPLE

PROBLEMS

A light plane has 10 m effective wingspan and 1.8mchord (regardless or airfoil chosen). It was originally designed to use a conventional (NACA 23015) airfoil section. With this airfoil, its cruising speedon a standard day near sea level is 225 km/hr. Aconversion to a laminar flow (NACA 662-215) section airfoil is proposed. Determine cruising speed that could be achieved for the same power.

A light plane has 10 m effective wingspan and 1.8m chord (regardless or airfoil chosen). It was originally designed to use aconventional (NACA 23015) airfoil section. With this airfoil, itscruising speed on a standard day near sea level is 225 km/hr. Aconversion to a laminar flow (NACA 662-215) section airfoil isproposed. Determine cruising speed that could be achieved for thesame power.

{FDV}23015 = P p = {FDV}66-215

FD = CD ½ V2A

assume efficiency same

{CD ½ V3A}23015 = {CD ½ V3A}66-215

CD = CD, + CD,i = CD, + CL2/(ar)

CD , CL for airfoil

for plane need CD,0

Assume airfoils should operate near design liftcoefficients.

(~0.3/47.6)(~0.2/59.5)

23015

662-215

CD = CD, + CD,i = CD, + CL2/(ar)

{~28% increase}

{CD ½ V3A}23015 = {CD ½ V3A}66-215

V66-215 = VD23015 (CD23015/CD66-215)

EXAMPLE

PROBLEMS

Aircraft with gross mass, m=4500 kg, flown in a circular path of 1 km radius at 250 kph. The plane has a NACA 23015With ar = 7 and lift area = 22 m2.

Find: Power to maintain level flight. P = FDV

Fig from 9.151

R = 1km

P = FDV

FD = CD (1/2V2Ap)

CD = CD, + CD,i = CD, + CL2/(ar)

Determine from force balance.Once know CL, can find CD, from Fig. 9-19

CL = FL / (1/2V2Ap)

FL = mg / cos()

FL cos ()

mV2/R

R = 1 km

FL sin ()

Fy = FLcos() – mg = 0Fr = -FLsin() = mar = -mV2/RFLsin() / FLcos() = (mV2/R) / mgtan () = V2/(Rg); = 26.2o

FL = mg / cos() = 49.2 kN CL = FL / (1/2V2Ap) = 0.754

Don’t know if flying at design CL, (and corresponding CD)but know weight and speed so can figure out CL, then find CD from graph.

CD = CD, + CL2/(ar)

CD ~ 0.007 for CL = 0.754 from Fig 9.19

CL = 0.754

CD = 0.007

CD = CD, + CD,i = CD, + CL2/(ar)

CD ~ 0.007 for CL = 0.754 from Fig 9.19

CD = 0.007 + (0.754)2/(7) = 0.0329

FD = FLCD/CL = 49.2kN x 0.0329 / 0.754 = 2.15kN

Power = FD V = 2.15kN x 250[km/hr] [1000(m/km)/3600(s/hr)]= 149 kW

W = FL = CL1/2 V2A; Vmin occurs for CLmax;

Vmin = [2W/ CLmaxA]1/2

FLAPS

W = FL = CL1/2 V2A; Vmin occurs for CLmax; Vmin = [2W/ CLmaxA]1/2

increase A to reduce Vmin: Vmin Vstall

= 0 = 0

= 0 = 15

TRAILING EDGE FLAPS-VARIES CAMBER

LEADING EDGE SLATS-POSTPONES STALL = 10o

= 30o+

= 25o

= 30o-

LEADING EDGE SLATS-POSTPONES STALL

Stall at 15o+

without leading edge slats

Not stalling yetwith leading edge slats

25% of c

40% of c

Maximum Lift: = 20o CL ~ 4 – 4.5

EXAMPLE

PROBLEMS

9.143

Airplane with effective lift area of 25 m2 is fitted with airfoils of NACA 23012 Section – conf. 2 (Fig. 9.23). Neglecting added lift due to ground effects determine the maximum mass of airplane if takeoff speed is 150 km/hr?

CL = FL/(1/2 V2Ap)

WFig. 9.2325 m21.23 kg/m3

Assume CL at lift off is CL max.

CL = 2.67; CL (1/2 V2Ap) = Wm = CL (1/2 V2Ap)/g = 7260 kg

NACA 23012

INTERESTING

FIGURE

1st jetliner

(1903, 30mph)

x

EXAMPLE

PROBLEMS

Lift Problem Examples – Relevant Equations

CL = FL/( ½ V2 Ap)CD = FD/( ½ V2 Ap)

Ap = max projection of wing

CL and CD values from wind tunnels are for 2-D airfoils

CD = CD, + CD,I = CD, + CL2/(ar)

CD = CD,0 + CD,I = CD,0 + CL2/(ar)

ar = b2/Ap

If steady flight: T = D and W = L = CL ½ V2Ap

CD for finite wing

Ex. 9.8: Given: W=150,000lbf, A=1600ft2, ar=6.5, CD,0=0.0182, =.00238 slug/ft2, Vstall=175mph, M0.6, c=759mph; steady level flight

Find optimum cruise speed.

(1) FD = CD ( ½ V2 Ap)(2) CD = CD,0 + CL

2/(ar)(3) CL = W/( ½ V2 Ap)

Optimum cruise speed = speed when FD/V vs V is minimum.

Use eq 3 to plug CL into eq 2, then plug CD from eq 2 into eq 1 Plot FD/V as a function of V between 175-455 mph (stall – 0.6 x c)and find peak.

optimum cruise speed

0

10

20

30

40

50

60

70

0 100 200 300 400 500 600

velocity (mph)

dra

g/v

elo

city

0

5000

10000

15000

20000

25000

0 100 200 300 400 500 600

level flight speed (mph)

Th

rust

= D

rag

(lb

f)

~ 325mph for optimum cruising

Ex. 9.8

GIVEN: W = 150,000 lbs; A = 1600 ft2; ar = 6.5; CD,0 = 0.0182; Vstall = 175 mph

FIND: (a) Drag from 175 mph to M = 0.6 (b) optimum cruise speed at sea level (c) Vstall and optimum cruise speed at

30,000 ft altitude

(a) Drag from 175 mph to M = 0.6

FDRAG = CD A (1/2) V2

CD = CD,0 + CL2/(ar)

CL = W/(1/2 V2)

150,000 lbf

0.00238 slug/ft3

175,…., 455 mph (M=0.6)

6.50.0182

1600 ft2 0.00238 slug/ft3

175,…., 455 mph (M=0.6)

Aircraft Characteristics

0

5

10

15

20

0 100 200 300 400 500

Speed V (mph)

Dra

g FD

(10

00 lb

f)

0

5

10

15

20

25

30

Po

we

r P

(10

00 h

p)

Drag force (1000 lbf)

Optimum line

P ower (1000 hp)

Optimum cruise speed at sea level, minimize FD/V

V (mph) 175 200 225 250 275 300 325 350 375 400 425 450 455CL 1.195874 0.915591 0.72343 0.585978 0.484279 0.406929 0.346733 0.298968 0.260435 0.228898 0.202761 0.180857 0.176904CD 0.088234 0.059253 0.043829 0.035015 0.029685 0.026309 0.024087 0.022577 0.021522 0.020766 0.020213 0.019802 0.019733FD (1000 lbf)11.06728 9.707257 9.087726 8.963245 9.19457 9.697927 10.42047 11.3275 12.39552 13.60812 14.95355 16.42327 16.73153P (1000 hp) 5.164729 5.177204 5.452635 5.975497 6.742685 7.758342 9.03107 10.57234 12.39552 14.51533 16.94736 19.70792 20.30093

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 50 100 150 200 250 300 350 400 450 500

velocity (mph)

Drag

/ Ve

loci

ty 323 mph

(b) optimum cruise speed at sea level

(c) optimum cruise and stall speed at 30,000 ft

FLIFT = W = CL A (1/2) SL VSL2

FLIFT = W = CL A (1/2) 30,000 V30,000

2

V30,000/VSL = [SL/ 30,000]1/2 = 1.63

V30,000 stall = 1.63 VSL stall; V30,000 op. cr. = 1.63 VSL op. cr.

THE

END

Lift force acting on an airfoil section can be evaluated using circulation theory (Kutta-1902;Joukowski-1906)

For an ideal fluid with no viscosityand a thin uncambered airfoil ofchord length c : Lper unit span = U

=circulation (Eq. 5-17; V•ds) = Uc[sin()]* Uc for small = density of fluidU = velocity of uniform flowL = U2c CL = U2c/(½ U2c) = 2

If no camber thenL = 0 at = 0

In ideal fluid slope = 2, viscosity reduces slope

separation

separation

*Proving this is beyond our scope but can be found in Anderson’s book: Fundamentals of Aerodynamics, pg 272

ASIDE