Transcript Chapter 9 F – Aug 23 (lift)

LIFT
LIFT
For most objects moving through
a fluid, the significant fluid force is drag.
CL
CD
However for some specially shaped
objects the lift force is also important.
LIFT - Some preliminaries:
????????????????????????????????????????????
LIFT

psin
p

Uo
  cos
w
L = dFy = -psindA + wcosdA
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
Lift force is the component
of R that is perpendicular to
free stream velocity, and
drag is the component of R
parallel to the free stream
velocity. If planes height is
not changing then:
Lift = Weight
Forces on airplane at
level speed and constant
height and speed.
Note: FL is not
parallel to N
FL
FD
Aplanform = chord x width of wing
FL
FD
CL(,Rec) = FL/
1
(/
2
2 V Ap)
CD (,Rec) = FD /(1/2  V2Ap)
Force generated if
we brought fluid
directly approaching
area to rest
CL =
FL/(1/2  V2Ap)
CD =
FD/(1/2  V2Ap)
Ap = planform area
max. proj. of wing
AP
Ap and c are
independent of 
AP
CD for most bodies (other than airfoils,
hydrofoils, vanes) is usually based on
the frontal area.
LIFT: Example – flat plate
Given: Kite in standard air, mass = 0.2 kg;
CL = 2sin(); CL/CD = 4. Find 
Area = 1 m2
5o
U= 10 m/s
=?
0.2kg (g)
CL = FL/(1/2 V2Ap)
CD = FD /(1/2  V2Ap)
Fy = FL – mg –Tsin() = 0
Fx = FD –Tcos() = 0
FL
Uo
5o
FD
mg
=?
CL = FL/(1/2 V2Ap)
Know: mass = 0.2 kg;
CL = 2sin(); CL/CD = 4;
Uo = 10 m/s; Find 
CD = FD /(1/2 V2Ap)
Fy = FL – mg –Tsin() = 0
*
*
Fx = FD –Tcos() = 0
*
FL
Uo
5o
FD
mg
=?
FL
U
FD
mg
T
=?
tan () = Tsin()/Tcos()
= (FL – mg)/FD
 = tan-1{{(FL – mg)/FD}
Know: Area, U, , mg, CL, CL/CD
9.144
FL
U
FD
mg
T
=?
Fy = FL – mg –Tsin() = 0
FL = CL A ½  U2
CL = 2sin(5o) = 0.548
FL = 33.7N
mg = 0.2(9.8)N
Tsin () = 33.7N – 0.2(9.8)N
9.144
FL
U
FD
mg
T
=?
Fx = FD –Tcos() = 0
FD = FL/4 = 8.43N
Tcos() = 8.43 N
FL
U
FD
mg
T
=?
tan () = Tsin()/Tcos()
 = tan-1{{(FL – mg)/FD} = 75.1o
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
Area
Force normal to plate, F = dp/dt
Time 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 sin2
Aside
Right Answer: CL = 2sin()
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
IMPORTANCE OF CAMBER
Ap
b
flat plate
bent plate
airfoil
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.
If camber (mean) line and chord line do not
overlap, then airfoil is cambered.
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.
Otto Lilienthal on a monoplane
glider in 1893
Otto Lilienthal on a biplane
glider in 1893
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,  = 0
NO LIFT
Inviscid flow,  > 0
NO LIFT
Inviscid flow,  > 0
+ circulation = LIFT
Lper unit span = U
= C vds = ro2ro
POTENTIAL FLOW
Lper unit span = U
= C vds = ro2ro
Lift?
Drag?
U = 4 m/s
R = 7.7 cm
Re = 4 x 104
=0
U = 4 m/s
R = 7.7 cm
Re = 4 x 104
 = 4U/R
Both develop lift,
see streamlines
pinched on top (faster
speeds, lower pressure)
and wider on bottom
(lower speeds and
higher pressure)
A CONSEQUENCE OF CIRCULATION
AROUND WING IS STARTING VORTEX
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.
LIFT = U
U

“Trailing vortices can be very strong
and persistent, possibly being a hazard
to other aircraft for 5 to 10 miles behind
a large plane – air speeds of greater than
200 miles have been measured.”
U = 30 cm/s
Chord = 180 mm
Re = 5 x 105
Floating tracer method
Starting vortex
(Munson also,
Fig. 9.38)
Both figures claim lift,
which figure’s streamlines are consistent with lift?
Lift & Bernoulli’s Equation
Physics 101 – Energy
Bernoulli’s Equation via Cons. Of Energy
Steady & Inviscid & Incompressible
Net work on fluid element when moved through stream tube:
Work = Increase in Mechanical Energy
Bernoulli’s Equation via Cons. Of Energy
Steady & Inviscid
Work = p1A1l1 - p2A2l2
Increase in Mechanical Energy =
[1/2  V22 + gz2]dVol - [1/2  V12 + gz1]dVol
p2 – p1 = 1/2  V22 + gz2 - 1/2  V12 - gz1
Lift & Bernoulli’s Equation
E
X
T
R
A
Momentum Eq.
BERNOULLI’S EQUATION
Via 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 dx
Y-MOMENTUM EQUATION: ½ d(v2) = - (1/) p/y dy
Z-MOMENTUM EQUATION:
½ d(w2) = - (1/) p/z dz - gdz
u 2 + v 2 + w 2 = V2
p/x dx + p/y dy + p/z dz = dp
½ d(V2) = - (1/) dp - gdz
BERNOULLI’S EQUATION
½ d(V2) = - (1/) dp - gdz
 {½  d(V2) = - dp - gdz}
INCOMPRESSIBLE:
½  (V22) - ½  (V12) = - (p2 – p1) -  g (z2 –z1)
p2 + ½  (V22) + z2 = p1 + ½  (V12) + z1
= constant along streamline
If irrotational each streamline has same constant.
BERNOULLI’S EQUATION
Momentum equation and steady, inviscid
and incompressible along a streamline.
p2 + ½  (V22) + gz2 = p1 + ½  (V12) + gz1
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
Momentum equation and unsteady, inviscid
and compressible along a streamline.
21V/ t ds + 21 dp/ + ½ (V22 – V12) + g(z2 – z1) = 0
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 tubes
so book not wrong.
? Turbulent flow?
? Turbulent flow?
From Fluid Mechanics
By Frank White
LIFT ‘Measurements’
Cp = (p-p)/(1/2 U2)
NOTE THESE ARE ‘MEASUREMENTS’ ON AIRFOILS (2-D)
= (p-p)/(1/2 U2)
2-D
Calculated (dots) and measured (circles)
pressure coefficients for airfoil at  = 7o.
favorable pressure gradient
unfavorable pressure gradient
= (p-p)/(1/2 U2)
2-D
*
*
As angle of attack increases
stagnation point moves
downstream along bottom
surface, causing an
unfavorable pressure gradient
at the nose*.
2-D
favorable
unfavorable
favorable to unfavorable
may cause lam. to turb. trans.
Stagnation Point
LIFT ‘Measurements’
CL = FL/( ½  V2Ap)
CD = FD /( ½  V2Ap)
NOTE THESE ARE ‘MEASUREMENTS’ ON AIRFOILS (2-D)
Rec = 9 x
106
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.
2-D
CL = FL/( ½  V2Ap)
Rec = 9 x 106
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
Re = 9 x 106
CL = FL/( ½  V2Ap)
Laminar flow sections
designed to fly at low
angles of attack, so
less drag but also
less maximum lift.
CD = FD/( ½  V2Ap)
2-D
Turbulent
Lam. – Turb.
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)
2-D
~ 15o
=
2-D
*
o
2
=
2-D
*
o
10
2-D
 = 15o - 
**
Check angle =15
separation at leading edge
 = 15o + 
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.
X
Plot is for one particular
Rec number
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
b
Ap
LIFT – DRAG POLAR
2-D
2-D
Higher the CL/CD the better!
LIFT – DRAG POLAR
CL proportional
to load
Note that x and y axis
Have different scales
CD related to drag that
plane must overcome
to achieve lift.
(does not include
fuselage drag, etc.)
Graph for one Re #
different angles
of attack
2-D
LIFT – Wings (3-D) vs. Airfoils (2-D)
Wing Tip Vortices
Wing tip vortices
(crop duster)
Schematic of subsonic
flow over the top of a
delta wing at an angle
of attack.
Two primary leading edge
vortices made visible by
air bubbles in water.
(Van Dyke Album of Fluid Motion)
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 tips
also 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) to
small air craft 5-10 miles behind large aircraft
The tendency for flow to leak around the wing tips
generally cause streamlines over the top surface of
the wing to veer to the wing root and streamlines
over 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 wing
without increasing the span.
The vortex which rotates
around from below the wing
strikes the winglet, generating
a 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 wings
will experience these effects more severely than a very long wing.
“New” glider by Wright brothers which was astoundingly successful
had an increase in wingspan to chord ratio from 3 to 6.
ar = b2/Ap
c
ar = b/c
Expect induced drag effects to scale with wing aspect ratio = b2/Ap
Soaring birds
- high aspect ratios
Maneuvering birds
- low aspect ratios
Pike
Bass
Tuna
Butterfly Fish
LIFT – Wings vs. Airfoils
Induced Drag
b2
b
p
p
“This causes the lift force to lean backwards a little, resulting in some
of the lift appearing as drag.” Fox et al.
V=V
eff is angle that wing sees between chord line and relative wind.
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 ar
must increase  ~ CL/( ar);
[linear]
airfoil
(2D)
wing
(3D)
wing
(3D)
For same lift as infinite ar,
CDi ~ CL  = CL2/(ar); [quadratic]
airfoil
(2D)
FOR INFINITE WING
CD, = FD/( ½ U2Ap)
CL, = FL /( ½ U2Ap)
FOR WING
CD = CD, + CD,i = CD, + CL2/( ar)
FOR AIRCRAFT
CD = CD,0 + CD,i = CD,0 + CL2/( ar)
CL ~ W /( ½ U2Ap) for steady state flight
PARTING NOTES
• 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)
• 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
TRAILING EDGE FLAPS-VARIES CAMBER
=0
=0
=0
 = 15
W = FL = CL1/2  V2A; Vmin occurs for CLmax; Vmin = [2W/  CLmaxA]1/2
increase A to reduce Vmin ; Vmin  Vstall
25% of c
40% of c
Maximum Lift:
 = 20o
CL ~ 4 – 4.5
LEADING EDGE SLATS-POSTPONES STALL
 = 10o
 = 25o
=
 = 30o+
30o-
LEADING EDGE SLATS-POSTPONES STALL
Not stalling yet
with leading edge slats
Stall at 15o+
without leading edge slats
25% of c
40% of c
E
X
A
M
P
L
E
P
R
O
B
L
E
M
S
A light plane has 10 m effective wingspan and 1.8m
chord (regardless or airfoil chosen). It was originally
designed to use a conventional (NACA 23015)
airfoil section. With this airfoil, its cruising speed
on a standard day near sea level is 225 km/hr. A
conversion 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 a
conventional (NACA 23015) airfoil section. With this airfoil, its
cruising speed on a standard day near sea level is 225 km/hr. A
conversion to a laminar flow (NACA 662-215) section airfoil is
proposed. Determine cruising speed that could be achieved for the
same power.
assume efficiency same
{FDV}23015 = P p = {FDV}66-215
FD = CD ½ 
2
VA
{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 lift
coefficients.
(~0.3/47.6)
(~0.2/59.5)
CD = CD, + CD,i = CD, + CL2/(ar)
23015
662-215
{CD ½  V3A}23015 = {CD ½  V3A}66-215
V66-215 = VD23015 (CD23015/CD66-215)1/3
{~28% increase}
E
X
A
M
P
L
E
P
R
O
B
L
E
M
S
Ex. 9.8: Given: W=150,000lbf, A=1600ft2, ar=6.5, CD,0=0.0182,
=.00238 slug/ft2, Vstall=175mph, M0.6, c=759mph; steady level flight
Find optimum cruise speed – Ex. 9.8
Optimum cruise speed = speed when FD/V vs V is minimum.
(1) FD = CD ( ½  V2 Ap)
(2) CD = CD,0 + CL2/(ar)
(3) CL = W/( ½  V2 Ap)
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.
Thrust = Drag(lbf)
25000
20000
15000
10000
5000
0
0
100
200
300
400
500
600
level flight speed (mph)
optimum cruise speed
70
drag/velocity
60
50
40
30
~ 325mph for
optimum cruising
20
10
0
0
100
200
300
velocity (mph)
400
500
600
E
X
A
M
P
L
E
P
R
O
B
L
E
M
S
Aircraft with gross mass, m=4500 kg,
flown in a circular path of 1 km radius
at 250 kph. The plane has a NACA 23015
With ar = 7 and lift area = 22 m2.
Find: Power to maintain level flight.
P = FDV
R = 1km
Fig from 9.151
FL F sin ()
L
FL sin () = mV2/R
mV2/R
R = 1 km
FL cos ()


W = FL cos ()
P = FDV
FD = CD (1/2V2Ap)
CD = CD, + CD,i = CD, + CL2/(ar)
CL = FL / (1/2V2Ap)
FL = mg / cos()
Determine  from force balance.
Once know CL, can find CD, from Fig. 9-19
Fy = FLcos() – mg = 0
Fr = -FLsin() = mar = -mV2/R
FLsin() / FLcos() = (mV2/R) / mg
tan () = V2/(Rg);  = 26.2o
FL = mg / cos() = 49.2 kN
CL = FL / (1/2V2Ap) = 0.754
CD = CD, + CL2/(ar)
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 ~ 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?
150 km/hr
CL = FL/(1/2  V2Ap)
Fig. 9.23
W
1.23 kg/m3
25 m2
NACA 23012
Assume CL at lift off is CL max.
CL = 2.67; CL (1/2  V2Ap) = W
m = CL (1/2  V2Ap)/g = 7260 kg
1st jetliner
x
(1903, 30mph)
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
1600 ft2
0.00238 slug/ft3
FDRAG = CD A (1/2)  V2
175,…., 455 mph (M=0.6)
CD = CD,0 + CL2/(ar)
0.0182
CL = W/(1/2  V2)
150,000 lbf
6.5
175,…., 455 mph (M=0.6)
0.00238 slug/ft3
FD = W {CD / CL} = W {FD/ [1/2  V2]} /{W/[1/2  V2]}
200
0.915591
0.059253
9.707257
5.177204
225
0.72343
0.043829
9.087726
5.452635
250
0.585978
0.035015
8.963245
5.975497
275
0.484279
0.029685
9.19457
6.742685
300
0.406929
0.026309
9.697927
7.758342
325
0.346733
0.024087
10.42047
9.03107
350
0.298968
0.022577
11.3275
10.57234
375
0.260435
0.021522
12.39552
12.39552
400
0.228898
0.020766
13.60812
14.51533
425
0.202761
0.020213
14.95355
16.94736
Aircraft Characteristics
20
30
15
Optimum line
P o wer (1000 hp)
20
10
15
10
5
Power P (1000 hp)
25
Drag fo rce (1000 lbf)
Drag FD (1000 lbf)
V (mph)
175
CL
1.195874
CD
0.088234
FD (1000 lbf)11.06728
P (1000 hp) 5.164729
5
0
0
0
100
200
300
400
500
Speed V (mph)
Optimum cruise speed at sea level, minimize FD/V
450
0.180857
0.019802
16.42327
19.70792
455
0.176904
0.019733
16.73153
20.30093
(b) optimum cruise speed at sea level
0.07
Drag / Velocity
0.06
0.05
0.04
323 mph
0.03
0.02
0.01
0
0
50
100
150
200
250
300
velocity (mph)
350
400
450
500
(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 V30,0002
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 viscosity
and a thin uncambered airfoil of
chord length c :
Lper unit span =  U 
ASIDE
separation
In ideal fluid slope
= 2, viscosity
reduces slope
 =circulation (Eq. 5-17; V•ds)
= Uc[sin()]*  Uc for small 
 = density of fluid
U = velocity of uniform flow
L =  U2c
If no camber then
L = 0 at  = 0
CL =  U2c/(½ U2c) = 2
separation
*Proving this is beyond our scope but can be found in
Anderson’s book: Fundamentals of Aerodynamics, pg 272
Lift Problem Examples – Relevant Equations
Re = 20,000
Angle of att
Symmetric
16% thick
CL and CD values from wind tunnels are for 2-D airfoils
CL = FL/( ½  V2 Ap) A = max projection of wing
p
2
CD = FD/( ½  V Ap)
CD for finite wing
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
Ex. 9.8: Given: W=150,000lbf, A=1600ft2, ar=6.5, CD,0=0.0182,
=.00238 slug/ft2, Vstall=175mph, M0.6, c=759mph; steady level flight
Find optimum cruise speed.
Optimum cruise speed = speed when FD/V vs V is minimum.
(1) FD = CD ( ½  V2 Ap)
(2) CD = CD,0 + CL2/(ar)
(3) CL = W/( ½  V2 Ap)
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.
Thrust = Drag(lbf)
25000
20000
15000
10000
5000
0
0
100
200
300
400
500
600
level flight speed (mph)
optimum cruise speed
70
drag/velocity
60
50
40
30
~ 325mph for
optimum cruising
20
10
0
0
100
200
300
velocity (mph)
400
500
600
Aircraft with gross mass, m=4500 kg,
flown in a circular path of 1 km radius
at 250 kph. The plane has a NACA 23015
With ar = 7 and lift area = 22 m2.
Find: Power to maintain level flight.
P = FDV
R = 1km
Fig from 9.151
P = FDV
FD = CD (1/2V2Ap)
CD = CD, + CD,i = CD, + CL2/(ar)
CL = FL / (1/2V2Ap)
FL = mg / cos()
Determine  from force balance.
Once know CL, can find CD, from Fig. 9-19
FL sin ()
FL cos ()
mV2/R
R = 1 km
Fy = FLcos() – mg = 0
Fr = -FLsin() = mar = -mV2/R
FLsin() / FLcos() = (mV2/R) / mg
tan () = V2/(Rg);  = 26.2o
FL = mg / cos() = 49.2 kN
CL = FL / (1/2V2Ap) = 0.754
CD = CD, + CL2/(ar)
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 ~ 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