LIFT

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

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

LIFTCL CD

LIFT - Some preliminaries:

????????????????????????????????????????????

LIFT

L = dFy = -psindA + wcosdA

psin

p

wcos

Uo

LIFT

suction

pressure

An object moving horizontally will experience lift if that object produces an asymmetrical (top-to-bottom) flow field.

LIFT

DRAG

Note: Lift is defined as the force that is perpendicular to the incoming flow and drag parallel to the incoming flow

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

FL

FD

Note: FL is not parallel to N

Aplanform = chord x width of wing

CL(,Rec) = FL/(1/2 V2Ap)

CD (,Rec) = FD /(1/2 V2Ap)

FL

FD

Force generated if we brought fluid

directly approaching area to rest

Ap and c are independent of

CL =FL/(1/2 V

2Ap)

CD =FD/(1/2 V

2Ap)

Ap = planform areamax. proj. of wing

CD for most bodies (other than airfoils, hydrofoils, vanes) is usually based on

the frontal area.

AP

AP

LIFT: Example – flat plate

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

= ?

CL = FL/(1/2 V2Ap)

CD = FD /(1/2 V2Ap)

Fy = FL – mg –Tsin() = 0

Fx = FD –Tcos() = 0

FD

mg

= ?

FL

Uo 5o

CL = FL/(1/2 V2Ap)

CD = FD /(1/2 V2Ap)

Fy = FL – mg –Tsin() = 0

Fx = FD –Tcos() = 0

FD

mg

= ?

FL

Uo 5o

Know: mass = 0.2 kg; CL = 2sin(); CL/CD = 4;

Uo = 10 m/s; Find

* *

*

tan () = Tsin()/Tcos() = (FL – mg)/FD

= tan-1{{(FL – mg)/FD}

FL

U FD

T

mg

= ?

Know: Area, U, , mg, CL, CL/CD

9.144

Fy = FL – mg –Tsin() = 0FL = CL A ½ U

2

CL = 2sin(5o) = 0.548FL = 33.7Nmg = 0.2(9.8)NTsin () = 33.7N – 0.2(9.8)N

FL

U FD

T

mg

= ?

9.144

Fx = FD –Tcos() = 0FD = FL/4 = 8.43NTcos() = 8.43 N

FL

U FD

T

mg

= ?

tan () = Tsin()/Tcos() = tan-1{{(FL – mg)/FD} = 75.1o

FL

U FD

T

mg

= ?

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, F = dp/dtTime rate of change of the normal component of momentum =(mass flow) x change in normal component of velocity =

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

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

Aside

Area

Right Answer: CL = 2sin()

Benjamin Robins (1707 – 1751) invented whirling arm for measuring aerodynamic forces. Borda in 1763

experimentally showed that lift on a plate varies as U2

sin and not U2 sin 2 as Newton suggested.

First wind tunnel built in 1884 by Horatio Phillips

LIFT: Camber

For all cases angle of attack is 4o and aspect ratio (b2/Ap) is 6.Lift to drag ratios of about 20 are common for modern transport planes.

flat plate bent plate airfoil

IMPORTANCE OF CAMBERAp

b

If camber (mean) line and chord line do not overlap, then airfoil is cambered.

Otto Lilienthal on a monoplaneglider in 1893

Otto Lilienthal on a biplaneglider in 1893

Otto Lilienthal (1848-96) is universally recognized as the first flying human. His wings were curved.On August 9th, 1896 Lilienthal suffered a fatal spinal injury, falling 10-15 meters from the sky.

why do airplanes fly?

Lift = U

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

PG 448 OUR BOOK

Inviscid flow, = 0NO LIFT

Inviscid flow, > 0NO LIFT

Inviscid flow, > 0+ circulation = LIFT

Lper unit span = U

= C vds = ro2ro

POTENTIAL FLOW

Lper unit span = U= C vds = ro2ro

Lift?Drag?

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

LIFT = U

U

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.”

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

(Munson also,Fig. 9.38)

Lift & Bernoulli’s Equation

Physics 101 – Energy

Net work on fluid element when moved through stream tube: Work = Increase in Mechanical Energy

Bernoulli’s Equation via Cons. Of Energy

Steady & Inviscid & Incompressible

Work = p1A1l1 - p2A2l2

Increase in Mechanical Energy =

[1/2 V22 + gz2]dVol - [1/2 V1

2 + gz1]dVol

p2 – p1 = 1/2 V22 + gz2 - 1/2 V1

2 - gz1

Bernoulli’s Equation via Cons. Of Energy

Steady & Inviscid

Lift & Bernoulli’s Equation

Momentum Eq.

EXTRA

BERNOULLI’S EQUATIONVia Momentum Eq.

X-MOMENTUM EQUATION:INVISCID:

(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]

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

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

BERNOULLI’S EQUATION

X-MOMENTUM EQUATION: ½ d(u2) = - (1/) p/x dxY-MOMENTUM EQUATION: ½ d(v2) = - (1/) p/y dy

Z-MOMENTUM EQUATION: ½ d(w2) = - (1/) p/z dz - gdz

½ d(V2) = - (1/) dp - gdz

u2 + v2 + w2 = V2

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

BERNOULLI’S EQUATION

½ d(V2) = - (1/) dp - gdz {½ d(V2) = - dp - gdz}

INCOMPRESSIBLE:

½ (V22) - ½ (V1

2) = - (p2 – p1) - g (z2 –z1)p2 + ½ (V2

2) + z2 = p1 + ½ (V12) + z1

= constant along streamline

If irrotational each streamline has same constant.

BERNOULLI’S EQUATION

p2 + ½ (V22) + gz2 = p1 + ½ (V1

2) + gz1

Momentum equation and steady, inviscid and 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

BERNOULLI’S EQUATION

21V/ t ds + 2

1 dp/ + ½ (V22 – V1

2) + g(z2 – z1) = 0

Momentum equation and unsteady, inviscid and compressible along a streamline.

Can be shown – White / Fluid Mechanics 3rd Ed.

Pgs 156-158

Using Bernoulli’s Equation (or not)

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

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

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

WHAT IS WRONG WITH THIS PICTURE?

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

BERNOULLI EQUATION, B.E., GOOD FOR STREAM TUBES WHERE ENERGY IS NOT BEING ADDED OR SUBTRACTED

Yet one can argue that B.E. is valid

for outer stream tubesso book not wrong.

From Fluid Mechanics

By Frank White

? Turbulent flow? ? Turbulent flow?

LIFT ‘Measurements’

NOTE THESE ARE ‘MEASUREMENTS’ ON AIRFOILS (2-D)

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

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

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

2-D

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

unfavorable pressure gradient

favorable pressure gradient

2-D

As angle of attack increasesstagnation point moves

downstream along bottom surface, causing an

unfavorable pressure gradient at the nose*.

*

*

2-D

unfavorable

favorable

favorable to unfavorablemay cause lam. to turb. trans.

Stagnation Point

LIFT ‘Measurements’

NOTE THESE ARE ‘MEASUREMENTS’ ON AIRFOILS (2-D)

CL = FL/( ½ V2Ap) CD = FD /( 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

2-D

Rec = 9 x 106

CL = FL/( ½ V2Ap)

Typical lift coefficient is of the order unity. Hence typical lift force is about equal to the product of the dynamic pressure times the planform area. FL ~ ½ V2Ap

Wing loading = FL/Ap

1903 Wright Flyer = 1.5 lb/ft2

Boeing 747 = 150 lb/ft2 bumble bee = 1 lb/ft2 2-D

Laminar flow sections designed to fly at low

angles of attack, soless drag but also

less maximum lift.

Re = 9 x 106

Turbulent Lam. – Turb.

CD = FD/( ½ V2Ap)

CL = FL/( ½ V2Ap)

2-D

LIFT - SEPARATION

Stall results from flow separation

over a major section of the upper surface

of airfoil

Rec = 9 x 106

CL = FL/( ½ V2Ap)

~ 15o2-D

= 2o

*

2-D

= 10o

*

2-D

*

= 15o -

= 15o +

separation at leading edge

Check angle =15*

2-D

Separation caused by unfavorable pressure gradient resulting from reduction in external flow.

LIFT – DRAG POLARS

Lift-Drag Polars are often used (Otto Lilienthal) to present airfoil data.

Plot is for one particular Rec number

X

Maximum L/D usually occurs at an angle of attack between 4° – 5° or where the CL is around 0.6.

Plot is for one particular Rec number

L/D ~ 400 for ar (b2/Ap) = L/D ~ 40 for sailplane with ar (b2/Ap) = 40

L/D ~ 20 for typical light plane with ar (b2/Ap) = 12

Ap

b

LIFT – DRAG POLAR

Higher the CL/CD the better!

2-D

2-D

CL proportional to load

CD related to drag thatplane must overcome

to achieve lift.(does not include fuselage drag, etc.)

LIFT – DRAG POLAR

Graph for one Re #different angles

of attack

Note that x and y axisHave different scales

2-D

LIFT – Wings (3-D) vs. Airfoils (2-D)

Wing Tip Vortices

Wing tip vortices

(crop duster)

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.

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.

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

Trailing vortices can be a hazard (200 mph) tosmall air craft 5-10 miles behind large aircraft

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.

Winglets work by increasing the effective area of the wingwithout increasing the span.

The vortex which rotates around from below the wing strikes the winglet, generatinga small lift force.

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

Soaring birds- high aspect ratios

Maneuvering birds- low aspect ratios

Tuna Butterfly Fish

Pike

Bass

LIFT – Wings vs. Airfoils

Induced Drag

p

p

b

b2

eff is angle that wing sees between chord line and 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 )CD,I ~ CL I ; i ~ CL/( ar) [theory/exp]

CD,I ~ CL2/( ar)

* Fundamentals of Aerodynamics by Anderson

*

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

For same lift as infinite ar, CDi ~ CL = CL

2/(ar); [quadratic]

airfoil (2D)

wing (3D)

wing (3D)

airfoil (2D)

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

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

2/( ar)

CL ~ W /( ½ U2Ap) for steady state flight

FOR WING

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

• 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!

• Induced drag can be as much as pressure and skin-friction drag (depends on speed)

PARTING NOTES

• Induced drag can be as much as pressure and skin-friction drag (depends on speed)

Components of the total drag of a modern airliner

HIGH-LIFT DEVICES

FLAPS

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

Vmin = [2W/ CLmaxA]1/2

= 0 = 0

= 0 = 15

TRAILING EDGE FLAPS-VARIES CAMBER

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

increase A to reduce Vmin ; Vmin Vstall

25% of c40% of c

Maximum Lift: = 20o CL ~ 4 – 4.5

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 c40% of c

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 airfoilfor 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)1/3

EXAMPLE

PROBLEMS

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 – Ex. 9.8

(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)

drag

/vel

ocity

0

5000

10000

15000

20000

25000

0 100 200 300 400 500 600

level flight speed (mph)

Thru

st =

Dra

g(lb

f)

~ 325mph for optimum cruising

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

FL cos ()

mV2/R

R = 1 km

FL sin ()

W = FL cos ()

FL sin () = mV2/R

FL

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()

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, which is 0.754, 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

?

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

150 km/hr

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

1st jetliner

(1903, 30mph)

x

Ex. 9.8GIVEN: 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)

FD = W {CD / CL} = W {FD/ [1/2 V2]} /{W/[1/2 V2]}

Aircraft Characteristics

0

5

10

15

20

0 100 200 300 400 500

Speed V (mph)

Dra

g F D

(100

0 lb

f)

0

5

10

15

20

25

30

Pow

er P

(100

0 hp

)Drag force (1000 lbf)Optimum linePower (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.

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

Lift Problem Examples – Relevant Equations

CL = FL/( ½ V2 Ap)CD = FD/( ½ V2 Ap)

Ap = max projection of wingCL 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)

drag

/vel

ocity

0

5000

10000

15000

20000

25000

0 100 200 300 400 500 600

level flight speed (mph)

Thru

st =

Dra

g(lb

f)

~ 325mph for optimum cruising

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

The End