Aeng 411 Notes Merged (2)
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aeng 411 notes merged (2)...
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AENG 411 Applied Subsonic Aerodynamics Aerodynamics
The MEAN AERODYNAMIC CHORD (MAC) is frequently used to define the moment coefficient.
WING THEORY
For straight tapered wings,
DEFINITION OF WING PROPERTIES
(Leave space for figure)
C = MAC =
Cr++ +
A simple construction to find the C on an arbitrary straight tapered planform is illustrated in the figure below.
(Leave space for figure)
(Leave space for figure)
–
GEOMETRIC WINGSPAN (b) is the distance between tip to tip of the wing, measured perpendicular to the airplane or wing centerline, regardless of the geometric shape of the wing.
–
WING AREA (S) is the projection of the plan form on a plane of reference which is usually the chord plane. WING ASPECT RATIO (A) may be defined as:
/
1. A = = the ratio of the square of the wingspan to the total wing area.
the average chord A = = the ratio of the total wing
2. A = = the ratio of the wingspan to 3.
area to the square of the average chord
TAPER RATIO λ – λ = /
defined as the ratio of the tip chord (ct ) to the root chord (cr)
Following the expression of force and moment equations for airfoils the wing aerodynamic forces & moment are written as:
= 12 = = 12 = = 12 = Rectangular Wing
= 2 SWEEP ANGLE Λ –
is the angle between the line perpendicular to the centerline and leading edge or the quarter chord line. It is denoted as
.
A straight, tapered wing of 30ft span has a leading-edge & trailing-edge sweep angle of , respectively. Its total area is 285.3 sq. ft. Find the magnitudes of root chord, tip chord & the mean aerodynamic chord.
45 & 15
= = 4.15′02′ =0.268 1 2 = 3 1
45
15 15 15 = 15 2
Sol’n
= 15..+.++ = 10.57
For ΔCAB Tan45 = = = = 15 For ΔCBD: tan15 = = tan15 = 4.02
CIRCULATION, DOWNWASH, LIFT AND INDUCED DRAG
–
Induced Drag part of the drag caused by lift
–
Downwash velocity the air velocity deflected perpendicular to the direction of motion of an airfoil, i.e. it is the bending down of the air column c olumn upon which the wing acts while in flight.
–
Angle of attack the acute angle between a reference line in a body and the line of the relative wind direction projected on a plane containing the reference line and parallel to the plane of symmetry.
–
Geometric Angle of Attack simply reffered to as the angle of attack, has been defined as the angle between the relative wind & wing cord Effective angle of attack - the angle of attack at which an airfoil produces a given lift coefficient in a two dimensional flow.
–
–
Induced Angle of Attack the difference between the actual angle of attack & the angle of attack for infinite aspect ratio of an airfoil for the same lift coefficient.
Absolute angle of attack the angle of attack of an airfoil, measured from the attitude of zero lift.
–
Critical Angle of attack the angle of attack at which the flow about an airfoil changes abruptly as shown by corresponding abrupt changes in the lift & drag.
–
Relative Wind the direction from which the air comes in meeting the wing, this direction being the direction of the airstream before it has been disturbed by the approaching wing. wing.
= ′
eq. 5
Since,
= = = =
eq. 6
Equating for L: (Leave space for figure)
Let:
S’ = swept area it is a cross
= downwash for behind the wing
-section) of the airstream taken perpendicular to the direction of mo motion tion of the wing.
= 1 = = 2
eq. 7
Prandtl has shown that, with semielliptic lift distribution, the swept area is a circle whose diameter is the span (b).
Two methods of calculating lift : (Leave space for figure)
1. Momentum method Force X Distance According to the linear momentum principle, assuming uniform
–
downwash over s’ = = ′
eq. 4
2. Energy method -the work done on the air mass per unit time equals the kinetic energy increase per unit time.
=′
From eq. 4
= =2 = But,
=
=
Equating for L:
= = = = For induced angle of attack,
=
, where
Where:
span efficiency factor or Oswald’s
e= efficiency factor (=0.85-0.95 for the wing alone)(= 1 for elliptical) Ae = effective aspect ratio TOTAL DRAG COEFFICIENT FOR A WING
= =
is in radian
For induced drag coefficient,
eqn. 1
Where is the (lift independent) sum of skin friction & pressure drag.
= = = = =
eqn. 2
Eqns 1 and 2 are valid for wings with elliptical loading (i.e., uniform. This condition can be achieved by using an elliptical planform. Note:
–
-For elliptical planform downwash is constant along the span -both the downwash and the induced angle of attack go to zero as the wing span becomes infinite.
Glavert has shown that for rectangular wings, more nearly correct formulas are:
A 3 4 5 6 7 8 9
= 1 = 1 τ 0.11 0.14 0.16 0.18 0.20 0.22 0.23
0.022 0.033 0.044 0.054 0.064 0.074 0.083
EXAMPLE # 1 A rectangular monoplane wing has a span of 14m & a chord of 2m. When determine: (a) induced angle of attack, (b) induced drag coefficient.
=0.42,
For non-elliptic lift distribution,
=
Given:
=14 =2 =0.42 Req’d: Sol’n: = 1 , = 18. 24 1 , . → 180⁄ = = 142 = 7 =0.2 =0.064 @ =7 421 0.20 = 18.240. 7 = 1.31 . (a) (b)
= 45 ℎ=3000 =0.85 = ? = 12 1 = 2
Req’d: Sol’n:
(a)
At level fight: (leave space for figure)
= = = = 12 = 1 2
From the table:
Subst. (b)
= 1 0. 4 2 = 7 10.064 =0.0085
12 1 = 2 = 1 2
EXAMPLE # 2: An airplane weighing 8500N, has a wing span of 12m. What is the induced drag at 3000m altitude if the airspeed is 45m/s? Assume e = 0.85.
But,
Given:
Subst.
= 8500 =12
= = 1 2
= 1 2
For wing # 2
= , = − Equating ,
= − = − = − = − = [ 1 − 1]
Where:
= . ℎ = 1 =1.225 . 0 0651 (−0. )3000 1 288 =0.908 8500 = 0.8510.90812 45 2 =204.37
Since,
= = − = = [ 1 − 1 ] = 1 [ 1 − 1 ] = 1 [ 1 − 1 ]
Multiply both sides by
AIRFOIL CHARACTERISTICS CORRECTION Assume that two wings have high but different aspect ratios. Also assume that they have the same airfoil. In that case, according
to Prandtl’s Lift Line Theory if these wings − ℎ ℎ = − ℎ are placed at the one effective angle of attack and , their lift coefficient
Where:
= # 1 = # 2 Let:
= = ∞∞ = 1 [ 1 − 1 ] ∞ ∞ = 1 ∞
the angle of zero lift.
For wing # 1
= , = −
Where:
∞
= slope of lift curve of wing with finite aspect ratio = slope of lift curve of wing with infinite aspect ratio
EXAMPLE: The slope for the lift coefficient curve for infinite aspect ratio is 0.09 per degree. What is the for a wing with an aspect ratio of 7 at an angle of attack of 9 degrees measured from an angle of zero lift? Assume = 0.85.
0.15 1.2 If this airfoil is used to construct an elliptical wing of , determine the wing lift curve slope.
= 7.0
Given: NACA 23012
Sol’n
−
Elliptical wing (
Given:
Req’d:
∞ = 0.09 = 7 = 9 . = ∞ = 1 ∞ ∗ ∞ = 1 ∞ ∞ = 1 18.24∞ 0.09 deg 9 = 18. 0.deg09 2 4 deg 1 70.85 =0.635
EXAMPLE Prob. 4.2 The test results of the NACA 23012 airfoil show the following.
Req’d
=1
0.15 1.2
)
@=7
(leave space for figure)
= ∆ − 1.2 − 0.15 ∞ = ∆ = − = 9 − 0 Sol’n:
∞ = 0.117 = 1∞∞ = 1 18.∞24∞ 0.117 = 18.24 deg0.117 1 7 = 0.090
EXAMPLE PROB. 4.3 A rectangular wing model of 40 in by 5 in has the ff. characteristics determined from a wind tunnel test:
=0.87, = = 0.09 deg& = −3,
if a full-scale rectangular wing of 42ft by 6ft is constructed with the same airfoil section, what lift will it develop at under standard sea-level conditions? Assume
= 5 & 120 ℎ =0.87 ℎ − . Given: Model
Full Scale
SSLC
=40. =5 = 0.8 09 = =0.87 = −3 = 42 = 6 = 7 = 5 =120ℎ =0.87 = ?
Sol’n = 12 = − = 118.24 [ 1 − 1 ] 0. 0 9 = 118.240.09 [ 1 − 1 ] 70.87 80.87 = 0.087
= 0.087 5 −−3 =0.696 = 0.696 0.002377 120∗ 426 =6,457.05 L
PART I. FUNDAMENTALS OF FLIGHT MECHANICS FOR STEADY SYMMETRICAL FLIGHT
Fig . Definition of angles and velocities in steady symmetrical flight Where: Xb, Y b, Zb – Body axes system (Yb, not shown, is pointing in the paper), with X b along some airplane reference line. Xs, Y s, Zs – Stability axes system (Y s is pointing along Yb) with Xs pointing in the direction of the velocity vector v.
γ – read (climb path angle). The flight path angle. Positive for ascending flight (climb) and negative for descending flight (glide or dive) α – airplane (airframe) angle of attack θ – pitch altitude angle v – true airspeed vh – horizontal flight speed component vv – vertical flight speed component or rate of climb (R.C.)
δ – rate of descent (R.D.)
Fig . Definition of forces in steady symmetrical flight.
+ ΣFxs = 0 D + Wsin γ - Tcos(α + θT) = 0 CDqS + Wsin γ - Tcos(α + θT) = 0 + ΣFzs = 0 L + Tsin(α + θT) – Wcos γ = 0 CLqS + Tcos(α + θT) - Wcos γ = 0 Where: L – airplane lift D – airplane drag T – airplane thrust W – airplane weight
θT – thrust orientation angle relative to body x-axis
Seven (7) quantities needed to completely define a steady symmetrical state: 1. W 2. h (through ρ) 3. α 4. θT 5. v 6. γ 7. T In most problems, W, θT and h will be given, leaving for variables. Two variables can be arbitrarily selected. Example cases are
Level flight ( γ=0) at speed υ. The variables T and α follow from the equations. Flight at a given thrust-level, T and a desired climb angle, γ. The variables v and α follow from the equations
̅ = ̅ + ̅ 11 = ̅ = [( + )̅ ] = 1 ̅ = ( + )̅ = 1 = ( + ) =̅ .2
Part 2 Unpowered Flight or Glide
R
Vv = V cos α Vv = RD = V sin α
β XB XS
W sin α ZS
Note:
>0 >0 >0 Fig. 3 Airplane in Gliding Flight
In this flight condition, T = 0
∑ = 0 − =0 =
= 0 −sin=0 =sin ̅ = sin .3 = 0 −cos=0 =cos ̅ = cos .4 ̅ = − ), therefore: Introducing sin̅.5 ̅ .6 ̅ = sin ̅ .7 ̅ = cos RD = V
But:
= − Substitute;
= + = 1= = + =
Note that the variables here are , h (through ), v & . By selecting
, ̅
& h, three variable remain of which one can be arbitrarily selected.
Part IV – Horizontal Distance Covered in a Steady Glide
Where: h= Initial altitude R= Horizontal distance covered in a steady glide
R
t=an̅= = tan= = =ℎ =ℎ ̅ t a n = tan̅= ⁄1 ̅ =tan−1 ⁄1 From the given figure
̅
Minimum Glide Path Angle γ min & Maximum C L/CD It is seen that to achieve the lowest possible , it is necessary to maximize C L/CD. For a parabolic drag polar CL/CD can be maximize by setting its derivative with CL to zero.
eqn. 1 eqn. 2
But,
. Then Substitute to eqn. 1
=0 = = − += 1− =0 + =0 =0 == = √ Substitute
eqn.3
Hence,
eqn. 4
*
From,
eqn. 5
== == 2 == 1√ = 1 Substitute
to the polar drag equation
Polar Drag Equation
Glide Path
For
Minimum Rate of Descent
= 1 =0 [ ] =0 += − ()− + + 311 4=0 3 3 =04 =0 3 =0 It is seen that to achieve the lowest possible RD, it is necessary to minimize & to maximize parabolic drag polar,
. For a
can be maximize by setting
its derivative with C L to zero.
*
3=3 =0 ==3√ 3 == == 4 3 = √ = √ 16 = 161 = 16 = 16 ,
(CdCl⃒) max= 12 πCdoAe .. ⃒max=
MINIMUM AIRSPEED OR STALLING SPEED
VMIN = VS =
WP
equation 17
=
33.71
To achieve a low minimum airspeed (stalling
w
Rmax = 1500 ft. (33.71)
speed), it is seen that should be low and C LMAX should be high.
Rmax = 50 565 ft.
EXAMPLE PROBLEMS 1. A glider weighs 800 lbs. and has a wing loading of 12 psf. Its drag equation is: C D = 0.010 + 0.022 C L2. After being launched at 1500 ft. in still air, find:
V @ Rmax =
Wρ √ πAeC ..
CL =
DO
= a) The greatest distance it can cover b) The greatest duration of flight possible over level ground In both cases, find the corresponding flight speeds. Ignore the effect of density changes of the atmosphere and use standard sea level conditions. Given: W = 800 lbs. W/s = 12 psf CDo = 0.010 πAe =
.
h = 1500 ft Required: a) RMAX & V @RMAX b) tMAX & V @ RDMIN
RMAX = h(CL/CD) MAX
V @ Rmax =
12 psf .77ug/.7
V @ Rmax = 122.39 ft/s
CcL ⃒max= 163 3πAE CD D clcd ⃒max= 163 (0.0122) 0.0220.3 010 m⃒ ax = ⃒ 12 psf .77ug/99. ⃒ 995.24
3
ρ = 0.002377 slug/ft (S.S.L.C.)
Solution:
CL = 0.674
RD min=
RD min= 3.19 ft/s
tmax =
RDin .9 / =
tmax =470.22 s
CL =
(CdCl⃒)max= 21.30
3πAeC ..
=
Rmax = h
CL = 1.17
= (21.30)(2000 ft)
Rmax = 42600 ft
wP 12 psf .77 .7
V @ RDmin =
γmin = tan-1
= tan-1
.
γmin = 2.69°
=
ρ @ h = 2000 ft
V @ RDmin = 92.90 ft/s
1+ T . −) ( [1+ 9 R ]
ρ=
2. What greatest horizontal distance can be traveled in a glide if the airplane weighs 4500 lbs, a rectangular wing 42 ft by 7 ft and glides from 2000 ft? the polar drag equation of the airfoil is C D = 0.0340 + 0.0162 CL2. Compute also the minimum glide path angle and minimum rate of descent. Given: W=4500 lbs S = b x c = 42 ft x 7 ft =294 ft2 h = 2000 ft CDO = 0.0340
4.26
=
4.26
Ρ= 2.24 x 10-3 slug/ft3
CcL ⃒max= 163 3πAE CD D cdcl ⃒max= 163 (0.01162) 0.01620. 3 034 cdcl ⃒max=854.18 Wρ/ 9 . −.
RDmin = πAe = 1/0.0162 Required: Rmax, γmin, RDmin Solution:
(CdCl⃒) max= 12 πCdoAe (CdCl⃒) max= 12 0.01620.1 0340
=
RDmin = 4.0 ft/s
3. Determine the minimum rate of descent of the airplane described in problem 3 if the the wing will be change into a straight-tapered with configuration as shown:
Given: W =4500 lbs S = trapezoid = ½ b (h 1 + h2) h = 2000 ft CD = 0.0340 + 0.0162CL2 CDO = 0.0340 ΠAe = 1/0.0162 Required: RDmin Solution S= =
b h1+h2 21 ft4 ft+8 ft
s = 126 ft2 x 2 s = 252 ft2 RDmin =
Wρ/ . −.
=
RDmin = 4.32 ft/s
Altitude, h(ft) Sea level 5,000 50,000
= = = ..
Density ratio, σ But,
Then,
STEADY POWERED FLIGHT For stationary (steady), symmetrical powered flight the equations of motion are:
where
= 0 = 0
= Thrust Orientation Angle and γ = flight path angle and
In the case of conventional airplanes it turns out that the
component to other forces in Tsin(α+θT) is small when compared to
Where: Tv = PAV = power available from the propulsive system Dv = PREQD = power required to overcome the drag at a given speed v WRC = climb power Therefore in steady symmetrical flight the power available equals the sum of the power required and the climb power
other forces in Tsin(α+θT) + L – Wcosγ For that reason it can be assumed that sin(α+θT)≈0 and cos(α+θT)=1.0. (Insert figure here) Hence equations above become,
= = = And
It is useful to write
multiplication by the airspeed V.
in terms of “work” by
In the case of level flight, RC = 0
+ ℎ = 0 ←∑ = 0
= = ∑ = 0 ==0 = =
The drag in level flight is
For a given weight, flight configuration and altitude
=
equations above the certain variables α, v, and t. One of these
variables can therefore be arbitrarily selected. Note that in this case can be written as:
= = = = = = =
Therefore, the power required can be written as
(excess power) (level flight)
Aerodynamic center – lift
Center of gravity – weight This level flight speed follow from equation
1 = 2 =
= (2)(1) = = = = = ( ) (2)(1) = ( ) (2) = ( ) (2)1⁄
= =
Report No. 6 Calculation of drag and power required for a low speed turboprop transport.
Drag and power required for the case of parabolic drag polars
Given: If a drag polar can be represented by Airplane weight Wing loading Altitude Aerodynamic characteristics Explanation of Table
W = 18,400 kgf
= 220⁄ h = 0m
= = 12 = 12 1 = 1 = 2 2 = 12 12 = 12 = = 12
The drag at some V is given by
Fig 8.17 Or
From:
Substitute:
1 1 = 2 2 12 = 12 12
Where:
Do = Parasite drag
Di = Induced drag The power required in level flight therefore follows that
= = 12 12 = = 12 12 = 12 12⁄ =
Since,
Where:
It is important to recognize that induced drag and induced power required (at 0 given altitude and Speed) is independent on the span loading. The lower the span loading, the lower the induced drag and induced power required.
Flight Speed at Minimum Drag
= 0 12 12 = 0 2 12 12 2− = 0 = 4 = 0 4 = 0 = 4 = ( ) (2) (1 ) = ( ) (2) √ 1 = ( ) (2)√ 1 = 12 12
= 12 [( ) (2)√ 1 ] 12 [ 2 1 ] √ = = 2 = = 2
1 2 1 [ ] 4 √ = 1 [ 2 1 ]⁄ 2 √ = 2 = 2 (1 )(2) = 2 ( ) (2)
POWER REQUIRED AT MINIMUM DRAG
= 12 12 ⁄ 1 2 1 = 2 [( ) () √ ] 1 [ 2 1 ] 2 √
HIGH SPEED MINIUM POWER REQUIRED
= 0 12 12 = 0 12 3 12 − = 0
32 = 21 2 3 = 4 = ( ) (2) (31 ) = ( ) (2)√ 31 = ( ) (2)√ 31
1 2 1 [ ] 4 3 √ = 1 [ 2 1 ]⁄ 2 √ 3 1 3 = [ ]⁄ 2 3 = 43 2 3 = 43 ( ) (2) 3
POWER MINIMUM REQUIRED
= 12 12 ⁄ 1 2 1 = 2 [( ) ()√ 3 ] 1 [ 2 1 ]⁄ 2 √ 3
DRAG AT MINIMUM POWER REQUIRED
= 12 12
= 12 [( ) (2)√ 31 ] 1 2 12 = 3 3
Altitude, h(ft) Sea level 5,000 50,000
= = = ..
Density ratio, σ But,
Then,
STEADY POWERED FLIGHT For stationary (steady), symmetrical powered flight the equations of motion are:
where
= 0 = 0
= Thrust Orientation Angle and γ = flight path angle and
In the case of conventional airplanes it turns out that the
component to other forces in Tsin(α+θT) is small when compared to
Where: Tv = PAV = power available from the propulsive system Dv = PREQD = power required to overcome the drag at a given speed v WRC = climb power Therefore in steady symmetrical flight the power available equals the sum of the power required and the climb power
other forces in Tsin(α+θT) + L – Wcosγ For that reason it can be assumed that sin(α+θT)≈0 and cos(α+θT)=1.0. (Insert figure here) Hence equations above become,
= = = And
It is useful to write
multiplication by the airspeed V.
in terms of “work” by
In the case of level flight, RC = 0
+ ℎ = 0 ←∑ = 0
= = ∑ = 0 ==0 = =
The drag in level flight is
For a given weight, flight configuration and altitude
=
equations above the certain variables α, v, and t. One of these
variables can therefore be arbitrarily selected. Note that in this case can be written as:
= = = = = = =
Therefore, the power required can be written as
(excess power) (level flight)
Aerodynamic center – lift
Center of gravity – weight This level flight speed follow from equation
1 = 2 =
= (2)(1) = = = = = ( ) (2)(1) = ( ) (2) = ( ) (2)1⁄
= =
Report No. 6 Calculation of drag and power required for a low speed turboprop transport.
Drag and power required for the case of parabolic drag polars
Given: If a drag polar can be represented by Airplane weight Wing loading Altitude Aerodynamic characteristics Explanation of Table
W = 18,400 kgf
= 220⁄ h = 0m
= = 12 = 12 1 = 1 = 2 2 = 12 12 = 12 = = 12
The drag at some V is given by
Fig 8.17 Or
From:
Substitute:
1 1 = 2 2 12 = 12 12
Where:
Do = Parasite drag
Di = Induced drag The power required in level flight therefore follows that
= = 12 12 = = 12 12 = 12 12⁄ =
Since,
Where:
It is important to recognize that induced drag and induced power required (at 0 given altitude and Speed) is independent on the span loading. The lower the span loading, the lower the induced drag and induced power required.
Flight Speed at Minimum Drag
= 0 12 12 = 0 2 12 12 22− = 0 = 4 = 0 4 = 0 = 4 = ( ) (2) (1 ) 1 = ( ) (2) √ 1 = ( ) (2) √ = 12 12
1] = 12 [[(( ) (2) √ 12 [[ 2 √ 1 ] = = 2 = = 2
1 2 1 [ ] 4 √ = 1 [[ 2 1 ]⁄ 2 √ = 2 = 2 (1 ) (2) = 2 2 ( ) (2)
POWER REQUIRED AT MINIMUM DRAG
= 12 12 ⁄ 1 2 1 = 2 [( ) () √ ] 1 [[ 2 1 ] 2 √
HIGH SPEED MINIUM POWER REQUIRED
= 0 12 12 = 0 12 33 12 − = 0
32 = 21 2 3 = 4 = ( ) (2) (31 ) = ( ) (2)√ 31 = ( ) (2)√ 31
1 2 1 [ ] 4 3 √ = 1 [ 2 1 ]⁄ 2 √ 3 1 3 = [ ]⁄ 2 3 = 43 2 3 = 43 ( ) (2) 3
POWER MINIMUM REQUIRED
= 12 12 ⁄ 1 2 1 = 2 [( ) ()√ 3 ] 1 [ 2 1 ]⁄ 2 √ 3
DRAG AT MINIMUM POWER REQUIRED
= 12 12
= 12 [( ) (2)√ 31 ] 1 2 12 = 3 3
CLIMB PERFORMANCE AND SPEED Equations of motion
Fig. Flight Path Geometry and Forces for Airplane in Accelerated Climbing Flight
Normal to flight path: +
∑ = 0 sin cos . ==0
. In normal (straight line) climbing flight, since = 0, the centrifugal force is zero. Assuming small climb angles < 15° and = 0. si=n0 cos . .= 0 where C.F. is the centrifugal force
=
Along the flight path:
→+ = 0
∝ sin ∙ = 0 sin ∙ = 0 sin = ∙ sin = 1 ∙ Since the rate of climb, R.C., is equal to the vertical component of the flight speed:
. .= ℎ = sin sin = .. .. = 1 ∙ ℎ ∙ ℎ .. = 1 ∙ ℎ ∙ . . . . = ∙ ℎ ∙ . . . .( 1 ℎ) = . .= 1 ℎ ℎ = Acceleration factor Where:
When the climb is made at a constant true airspeed,
. .= As written in its present form, eqn.
. .= −
⁄ℎ = 0
. Therefore:
is applicable to a jet plane. Since a
propeller-driven airplane is related in terms of power; instead of thrust, eqn.
− for propeller-driven airplane as: can bewritten
. . = . . = ′
. .=
. .= =
(Note: v is constant) CLIMB PERFORMANCE AND SPEED OF PROPELLER DRIVEN AIRPLANES POWER REQUIRED Power required for flight at constant speed is given by: Since for small γ,
And,
Therefore,
At sea level:
where level.
W o should
Equating
;
′ = ′ = = = ( ) (2)(1) ′ = 5501 ( ) (2)(1) 2 = 550 ( ) () = 550 ( ) (2)∙ = = ()(2)(1) ′ = 550 ()(2)∙
be interpreted as a reference weight and not as the weight at sea
= ′ = ′ ()
Equating
Equating
:
= ∙ = ()1 = ()(1)
: ′ = ′ () (1)
REPORT NO.7 POWER REQUIRED AT SEA LEVEL AND ALTITUDES PROBLEM: A single-engine ______________ aircraft has the following characteristics:
==116.165325 = 0. =0370. 1.65 064
(Clean Airplane)
Compute the power required at sea level, 5000ft, 1000ft, 15000ft, and 20000ft altitudes: FOR MINIMUM SPEED:
= ( ) (2)1 = (116. 165325 )(0.0023772⁄)(1.165) = 85.15 ⁄ × 1522 = 58.06 ℎ 22 = 12 = 1(0,002377 1653 ) (× 15 ) 116.25 2 3 ℎ = 5561.8 ℎ: . ℎ = (1 ) ==0. 003566 ° = 519° ℎ = Where:
TABLE 1 Altitude, h (ft) SEA LEVEL 5000 10000 15000
Density Ratio, 1.000
σ
20000
EXPLANATION OF TABLE 2 Column No.
①, ②, ③, ④, ⑤, ⑥, ⑦,
′ ′ ′
=
Given
=
.①
= = = = =
0.0370.064② ③② ④×① ①√ ⑤√
Power Available The available thrust power is given by:
= ×
Where: BHP = shaft brake horsepower
= propeller efficiency
REPORT NO. 8
POWER AVAILABLE AT SEA LEVEL AND ALTITUDES
PROBLEM: For the airplane described in Report No. 7, the engine rated at _______HP at ______RPM is installed. Find the data for the thrust horsepower available vs velocity curve. Given:
= = 116 = = 2800 = 78%
Solution: At level flight,
= = 1160.78 = × = 90.48
By interpolation: V0
THPREQ’Do
mph
hp
130
76.41
VDES
90.48
140
92.57
140130 130 = 90.4876.41 92. 5 776. 4 1 = 138.71 ℎ
POWER SPEED COEFFICIENT, C S
= ⁄ ==0.0 =02377 ⁄ __________ = = ________⁄ ∙ = = ___________
where:
In fig. 51: @Cs= _________ and ζ P=78%, BLADE ANGLE, β=_________ In fig. 52:
@β=_________ and Cs= ________,
=
Propeller Diameter, D
where:
= ⁄ == ==__________ ⁄ ________ =
Explanation of Table:
①,
V
=
Given
②,
% Design V
=
①
③,
% Design N
=
Obtained from Fig. 449
④,
N
=
③ x DESIGN N
⑤,
BHP
=
③ x DESIGN BHP
⑥,
=
①④D ⑥
⑦,
% DESIGN
=
⑧, ⑨, ⑩,
% DESIGN
=
Obtained from Fig. 55a
=
⑧ x DESIGN ⑤x⑨
=
⑪,
N
=
⑫,
BHP at reduced N
=
⑬,
BHP
=
⑭, ⑮, ⑯, ⑰, ⑱,
=
% DESIGN
=
% DESIGN
=
④x altitude× ⑪factor
⑫ x ALTITUDE FACTOR ①× ⑪×D ⑭ ÷Design
Obtained from
fig. 55a = =
(fig.55b)
⑯ x DESIGN ⑬x⑰
(fig. 44b)
CLIMB PERFORMANCE AND SPEED Equations of motion
Fig. Flight Path Geometry and Forces for Airplane in Accelerated Climbing Flight
Normal to flight path: +
∑ = 0 sin cos . ==0
. In normal (straight line) climbing flight, since = 0, the centrifugal force is zero. Assuming small climb angles < 15° and = 0. si=n0 cos . .= 0 where C.F. is the centrifugal force
=
Along the flight path:
→+ = 0
∝ sin ∙ = 0 sin ∙ = 0 sin = ∙ sin = 1 ∙ Since the rate of climb, R.C., is equal to the vertical component of the flight speed:
. .= ℎ = sin sin = .. .. = 1 ∙ ℎ ∙ ℎ .. = 1 ∙ ℎ ∙ . . . . = ∙ ∙ . . ℎ . .( 1 ℎ) = . .= 1 ℎ ℎ = Acceleration factor Where:
When the climb is made at a constant true airspeed,
. .= As written in its present form, eqn.
. .= −
⁄ℎ = 0
. Therefore:
is applicable to a jet plane. Since a
propeller-driven airplane is related in terms of power; instead of thrust, eqn.
− for propeller-driven airplane as: can bewritten
. . = . . = ′
. .=
. .= =
(Note: v is constant) CLIMB PERFORMANCE AND SPEED OF PROPELLER DRIVEN AIRPLANES POWER REQUIRED Power required for flight at constant speed is given by: Since for small γ,
And,
Therefore,
At sea level:
where level.
W o should
Equating
;
′ = ′ = = = ( ) (2)(1) ′ = 5501 ( ) (2)(1) 2 = 550 ( ) () = 550 ( ) (2)∙ = = ()(2)(1) ′ = 550 ()(2)∙
be interpreted as a reference weight and not as the weight at sea
= ′ = ′ ()
Equating
Equating
:
= ∙ = ()1 = ()(1)
: ′ = ′ () (1)
ACTIVITY NO.7 POWER REQUIRED AT SEA LEVEL AND ALTITUDES PROBLEM: A single-engine ______________ aircraft has the following characteristics:
==116.165325 = 0. =0370. 1.65 064
(Clean Airplane)
Compute the power required at sea level, 5000ft, 1000ft, 15000ft, and 20000ft altitudes: FOR MINIMUM SPEED:
= ( ) (2)1 = (116. 165325 )(0.0023772⁄)(1.165) = 85.15 ⁄ × 1522 = 58.06 ℎ 22 = 12 = 1(0,002377 1653 ) (× 15 ) 116.25 2 3 ℎ = 5561.8 ℎ: . ℎ = (1 ) ==0. 003566 ° = 519° ℎ =
Density Ratio from Sea Level up to Tropopause
Where:
TABLE 1 Density Ratio, Altitude, h (ft) SEA LEVEL
1.000
5000 10000 15000 20000
EXPLANATION OF TABLE 2 Column No. :
①, ②, ③, ④, ⑤, ⑥, ⑦,
′ ′ ′
=
Given
=
.①
= = = = =
0.0370.064② ③② ④×① ①√ ⑤√
Power Available The available thrust power is given by:
= ×
Where: BHP = shaft brake horsepower
= propeller efficiency
σ
REPORT NO. 8
POWER AVAILABLE AT SEA LEVEL AND ALTITUDES
PROBLEM: For the airplane described in Report No. 7, the engine rated at _______HP at ______RPM is installed. Find the data for the thrust horsepower available vs velocity curve. Given:
= = 116 = = 2800 = 78% 78%
Solution: At level flight,
== 1161160.0.788 = × = 90.48
By interpolation: V0
THPREQ’Do
mph
hp
130
76.41
VDES
90.48
140
92.57
140130 130 = 900..48 7676..41 92. 5 776. 4 1 = 138.71 ℎ POWER SPEED COEFFICIENT, C S
= ⁄ ==0.0 =02377 ⁄ __________ = = ________⁄ ∙∙ = = ___________
where:
In fig. 51: @Cs= _________ and ζP=78%, BLADE ANGLE, β=_________ In fig. 52:
@β=_________ and Cs= ________,
=
Propeller Diameter, D
where:
= ⁄ == ==__________ ⁄ ________ =
Explanation of Table Column No.:
①,
V
=
Given
②,
% Design V
=
①
③,
% Design N
=
Obtained from Fig. 44a
④,
N
=
③ x DESIGN N
⑤,
BHP
=
③ x DESIGN BHP
⑥,
=
①④D ⑥
⑦,
% DESIGN
=
⑧, ⑨, ⑩, ⑪,
% DESIGN
=
Obtained from Fig. 55a
=
⑧ x DESIGN ⑤x⑨
⑫,
BHP at reduced N
=
⑬,
BHP
=
⑭, ⑮, ⑯, ⑰, ⑱,
N
=
=
=
% DESIGN
=
% DESIGN
=
④x altitude× ⑪factor
⑫ x ALTITUDE FACTOR ①× ⑪×D ⑭ ÷Design
Obtained from
fig. 55a = =
(fig.55b)
⑯ x DESIGN ⑬x⑰
(fig. 44b)
ALTITUDE, h Ft 5000 10000 15000 20000
ALTITUDE FACTOR Fig. 44b Fig. E5b
CLIMB PERFORMANCE AND SPEED From the plotted graphs of the power required and available at given altitudes, various performance parameters can be obtained: Maximum flight speed Maximum rate of climb Maximum climb angle All corresponding speeds
maximum climb angle is not less than the stall speed. To find the condition for maximum (R.C. / v) drawn a straight line from the origin tangent to the curve.
ACTIVITY NO. 9
Climb Performance and Speed Problem: For the aircraft described in activity nos. 7 and 8, determine the maximum speed, the maximum R.C., the speed for maximum R.C., the best climb angle and the corresponding flight speed.
The maximum speed in level flight at a given altitude is simply the speed at which the power available and power required curves intersect. The maximum rate of climb occurs when the excess power is maximum. The maximum climb angle occurs when the ratio (R.C. /V) is at maximum
..= si .n. sin = = sin− (..) = sin− (..)
In determining the maximum climb angle, care should be taken to make sure that the speed for
h
= mph
hp
hp
hp
. 33,.=000 ft/min
600.00 ) m500.00 p f ( . 400.00 C . R , b 300.00 m i l c f 200.00 o e t a r 100.00
0.00 0
50
100
150
Flight speed, v (mph)
At h
sin− .. =
ALTITUDE, h
MAX R.C.
SPEED FOR MAX R.C.
FT
MPH
FT/MIN
MPH
SEA LEVEL 5,000 ...
BEST CLIMB ANGLE,
SPEED FOR BEST CLIMB ANGLE
DEG
MPH
CLIMB PERFORMANCE FOR CONSTANT THP AV AND PARABOLIC DRAG EQUATION For quick performance estimation, it is often assumed to be independent of flight speed. It is further assumed that the airplane drag equation is of the parabolic form:
= + ..= − = 550 ( ) (2)∗ ∝
For general aviation aircraft, typical values of a range from 0.7 to 0.85, Eqn.
Indicates
that maximum R.C. implies minimum PREQD. However, from eqn.
Therefore, the condition for
maximum R.C. requires
⁄
to be
maximum. That is the derivative of
⁄
With CL should be zero.
= 0 + = 0
2 ( + )(32 ) = 0 ( + ) 32 + 32 2 = 0 2 3 + 3 4 = 0 3 = 0 1 = 3 = 3 = 3 = √ 3 = + 3 = 4
An airplane weighs 36,000-lb has a wing area of 450-ft2. The drag polar equation is CD = 0.014 + 0.05 CL2. It is desired to equip this airplane with turboprop engines with available power such that a maximum speed of 602.6-mph at sea level can be reached. The available power is assumed to be independent of flight speed, calculate the maximum rate of climb and the speed at which it occurs.
Given: W = 36,000-lb S = 450-ft2 CD = 0.014 + 0.05 CL2 Vmax = 602.6-mph * SEA LEVEL
= 883.81fps
THρAV = constant
Req’d:
Max R.C. and Vmax R.C.
Solution:
Sea Level 200
) p h ( 150 P H T 100 , r e w 50 o P e s 0 r o 0 H t s u r h
At Vmax
Where: D = lb V = fps
THPAV = constant
50
100
150
Airspeed, V (mph)
= = 550
200
250
= 12 36, 000 = 12 0.002377 ⁄ 883.81⁄ 450 = 0.086 = 0.014+0.05 = 0.014+ 0.050.086 = 0.0144 = 12 = = (0.0.00144860)36000 = 6027.91 6027.95501883.81 = = = 9686.41 At minimum power required:
= √ 3 .. = (2)(1)
At maximum R.C.
= 1083. 0 3ℎ = √ 3 . . = 0 00 ×33, = 3 (0.105) 0.014 9686.436,11083. 0 3 = ( )33, 0 00 0 00 = 0.917 = 7886.42⁄ = 4 = 40.014 = 0.056 = = (0.0.095617) 36000 = 2,198.47 = 550 = 2198.45507270.93 = 1082.97ℎ CEILINGS AND TIME TO CLIMB
Rate of climb varies in proportions with altitude
= 0.917. = 15.68 0.056 = 550 (2) × 1 36000 2 1 = 36000 ( )( )× 550 450 0.002377 15.68
Using ratio and proportions:
.. = ...ℎ . = . ℎ.... .
straight
Using ratio and proportions:
.. = ..100 = ...100 .
Example: Given: W
=
1653lb
R.C.O =
588.33fpm
h
10,000ft
=
R.C.H=10000 =
127.97fpm
Where: H
=
absolute ceiling
Hs
=
service ceiling
h
=
any altitude
R.C.O =
rate of climb at sea level
R.C.HS=
rate of climb at service ceiling (=100fpm)
R.C.h =
rate of climb at any altitude
Absolute Ceiling, H
the maximum altitude above sea level at which a given airplane would be able to maintain horizontal flight under air conditions.
Service Ceiling, Hs the altitude above sea level; under air conditions, at which a given airplane is unable to climb faster than a small specified rate. -
-
-
The altitude where the rate of climb is 100 ft/min An airplane can steady, maintain 100ft/min For jets, 500ft/min
Required: H and Hs
Solution:
= . ℎ.... . ⁄ 3 3 = 588.10000588. 33⁄ 127.97 ⁄ = 12,779.78
∗∗ − 100 588.33⁄ − 100 12,779.78588.33 ⁄ 10,607.57 Example #2 Given:
4,000 @ 60ℎ @=, 17ℎ Required:
Solution: At sea level
.. ∗33,000 60 ∗ 33,000 .. 4000 .. 495/ At h = 10,000 ft
∗ 33,000 ∗ 17 ∗ 33,000 ∗ 4000 ∗ 140.25 / ∗ℎ∗ − ∗ ∗
10,000495 495 − 140.25 13,953.49 ∗∗ − 100 − 100 13,953.49495 495 11,134.60 Example #3 An airplane weighs 20kN; its rate of climb at sea level is 350 m/min, its absolute ceiling is 4km. What is its service ceiling? Given:
20 ∗ 350 / 4 Required:
Solution:
∗∗ −30.49 − 30.49 4 350 350 3.65
TIME TO CLIMB
Using ratio & proportion
ℎ ∗ ∗ − ∗ ∗ ∗ − ℎ ∗∗ ∗ ∗ ∗ − ℎ ∗ ∗ ∗ ∗ℎ − ℎ ∗ − ℎ ℎ ∫ ∗ ∫ − ℎ − ∗ ln − ℎ
− ∗ − ℎ − − − ∗ − ℎ − ∗ − − ℎ ∗ − ℎ Where: Example #1: At sea level, an airplane .et rate of climb is 350 m/min. Its absolute ceiling is 4.5 km. How long will it take to climb o 2 km? Given:
∗ 350 / 4.5 ℎ 2 Required:
Solution:
∗ − ℎ 4,500 )[( 4.5 )] (350 / 4.5 − 2 7.56 Example #2 At sea level, an airplane weighing 5,200 lb has rate of climb of 850 ft/min. Its absolute ceiling is 18,500 ft. (A)How long will it take to climb from SL to 15,000 ft? (B) How long will it take to climb from 15,000 ft to absolute ceiling?
(C)
How long will it take to climb for SL to H?
Given:
5,200 ∗ 850 / 18,500 Required:
=, (B) =, (C) (A)
Solution: (A)Time to climb from SL to h = 15,000 ft
∗ − ℎ 500 [( 18,500 )] 18, 350 / 18,500 − 15,000 36.24 (B) Time to climb from h = 15,000 ft to absolute ceiling At h = 15,000 ft
∗ ∗ − ℎ ∗∗ / ∗ 850 /min− 15,00018,350 500 ∗ 160.81 / ∗ − ℎ [( 18,500 )] 160.18,81500/ 18,500 − 3,500 24.13
(C)
Time to climb from SL to H
36.24 + 24.13 60.37 REPORT NO. 10 Ceilings and time to climb
①
Altitude, h ft Sea level 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 10,607.57 11,000 12,000 12,779.78
②
R.C. ft/min 588.33 634.37 680.40 726.44
③
Δh/R.C.
min 0
④ ③
t=Σ
min 0
346.77
127.97
0
ꝏ
Time to climb to service ceiling
∗ − 1.) A 22,420 N aircraft has an excess power of 56 kW at sea level and the service ceiling is 3.66 km. Determine: (a)Rate of climb at 2.3 km (b)Rate of climb at service ceiling (c) Rate of climb at absolute ceiling Given:
22,240 56 3.66 Required: a.) b.) c.)
.. .. ..
Solution:
56,000 .. 22,240 60 151.20 / .. 2.52 ∗ 1 For H,
.. ..−30.49 3660157.2/ .... −30.49 151.20 − 30.49 4584.09 .. ..− .. 151.20/ .. 151.20 − 28004584. 09 .. 58.85 / b.) .. 30.49 / c.) .. 0 a.)
2.) A light airplane has a service ceiling of 4km. Its rate of climb at sea level is 370 m/min. How long will it take to climb to 3 km and time to climb to reach service ceiling? Given:
5 .. 370
Required: a.) b.)
3
Solution:
.. ..−30.49 .... −30.49 3705370 − 30.49 5.45 a.)
.. − ℎ 5.45 ] (5450 )[ 370 5.45− 3 11.78 b.)
.. − 5.45 ] (5450 )[ 370 5.45 −5 37.43
TAKE-OFF AND LANDING PERFORMANCE The take-off and landing phases of flight are normally broken down as follows:
TAKE-OFF Accelerating ground-run Rotation Lift-off Climb-out
LANDING Descent Flare Touchdown Decelerating ground-run
For commercial airplanes, Fig 10.1 summarizes the take-off procedures to be followed in case of engine failure. The reader will observe that such a case a pilot has two options. 1. Continue take-off 2. Stop (Discontinue take-off) Item
Speeds
Climb
MILC5011A (Military)
Far Pt 23 (Civil)
Gear up 500fpm
Gear up 300 fpm
≥ 1.1 ≥ 1.2
FAR Pt 25 (Commercial)
≥ 1.1 ≥ 1.1 ≥ 1.2 ≥ 1.1 Geardown ½ % @ VLOF
@ S.L (AEO) 100 fpm @S.L (OEI) Take-off distance
Field Length Definition over 50’
@S.L (AEO)
Gear up 3% @ VCL (OEI)
Take-off 115% of takedistance off distance over 50’ with AEO over 35’ or
Notes:
balanced field length AEO = All Engines Operating
OEI = One Engine Inoperative
The operational ground rules governing the take-off of CTOL airplanes are summarized in Table 10.1. The operational speed called out in Fig. 10.1 and in Table 10.1 are important and have the following definitions: VS = One g stall speed out of ground effect VLOF = Lift off speed VR = Rotational speed, i.e., the speed at which rotation is initiated during that take-off to attain the V2 climb speed V1 = decision speed (option 1 or 2) V2 or VCL = climb out speed *CTOL – Conventional Take-off and Landing
Reverse Thrust Enroute Descent Bell out
Touch Down
Final approach
Initial approach
Flare
Fig. 10.2 Landing Procedures for a Multi-Engine Civil transport Airplane
Table 10.2 Summary of CTOL Landing Rules Item Speeds Field Length
MIL-C-5011A
≥ 1.3 ≥ 1.1
FAR Pt 23(Civil)
≥ 1.3 ≥ 1.15
FAR Pt 25(Commercial)
≥ 1.3 ≥ 1.15
Landing Distance
Landing Distance
Landing Distance
Over 50’
Over 50’
Over 50’ divided
by 0.6 Definition
Fig. 10.2 shows the landing produces normally followed for commercial airplanes. Table 10.2 summarizes the ground rules governing CTOL landing. The associated important speeds have the following definitions: VA = speed over the 50 ft obstacle (also called as the approach speed) VTD = speed at touchdown during landing It should be noted that all speeds specified in tables 10.1 and 10.2 are minimum speeds.
Part 6 Take-off Analysis
VCL R
VG
R
ӨCL
50
VLCF
STR
SG
SCL
SR
Fig 10.3 Geometry for Take-Off performance Analysis For analytical purposes, the take-off consists of ground run rotation, transition, and climb over a 50ft (35ft) obstacle. The total take-9off distance is therefore the sum of the ground distance (S G), Transition distance (STR) and climb distance (SCL). Fig. 10.3 illustrates the geometry used in the analysis of the take-off.
Report no.11 Take-off and Landing Performance
Ground distance, SG (General)
=± Where:
VG = Ground speed V = Airspeed VW = Wind Speed If VG = 0
=± = ± = ±
But,
=
(eq.1)
(eq.2)
Substitute (eq.2) in (eq.1)
= ±
(eq.3)
Integrating both sides
∫ = ∫ ̅ ±
± = ∫ ̅
(eq.4)
The upper sign in (eq.4) is for downwind take-off. The acceleration of an airplane during the ground run may be found by considering the forces acting on the airplane. The airplane is seen to be under the influence of lift, drag, thrust, weight and ground friction forces. The runway inclination produces two major effects:
∅
∅
(a)A force equal to W acting along the path to retard the motion if the airplane goes up the slope ( ) and (b)The normal force on the runway which is reduced and which yields a small decrease in the friction force.
∅
= = ∅ = =∅ = ∅ .5
± = ∫ ̅ ∅ Substitute (eq.5) in (eq.4)
± = ∫ ̅ ∅ Coefficient of relations , Concrete : 0.02-0.03
Hard turf : 0.05 Short grass : 0.05 Long grass : 0.10 Soft ground : 0.10-0.32
Wind effect on take-off
ℎℎ
=
-VW = 50% VW1
+VW = 150% VW1 Where: VW1 = wind speed at h 1 VW2 = wind speed at h 2 h1 = height of MAC from the ground h2 = height of tower (=50ft)
(eq.6)
Effect of speed on thrust
Sea level
Thrust
V A
VLDF
H T
Flight speed ,v (mph)
= = Where:
T = is in lbs. VLOF = is in mph
h=h1
MAC
Part 7 Effect of the ground on C L and CD
Lift coefficient in ground level effect
= ∝ ∝ ∝∆∝
Where: CL(OGE) = lift coefficient in the appropriate configuration out of ground effect
√
=
∝ +√ +/+ ∝ + + /+ =
=
In fig. 10.8
=ℎ
∆∝= . . ,. CR/2 CL
CT/2
Λ/
Drag Coefficient in Ground Effect
= ∆ = = ∆ = ′ , =1. 0 Where:
______+ ______
(clean airplane)
In fig 10.9
=ℎ For 0.033 < h/b < 0.25 , the ground influence coefficient from:
= .−.−.ℎ/ℎ/
′
can be estimated
Part 8 Approximate method
I
= −
for SG (eq.7)
Where:
=∅ =0 =∅ = Approximate for zero wind speed Assume VW = 0 , (eq.4) becomes
= ∫ 22 = ∫ 222 2 = ∫ 22 Let
= =2
(eq.8) (eq.9)
2 = ∫ 222 2 = 2 20 2 =2 [{22 2} 2 0 ] = [{2 2 2}
22]
= [ 2 {2}] 2 2 = 2 2 =
2 = 1
2 = = Where:
(eq.11)
(eq.10)
− =
(eq.12)
Example 10.1 The wing characteristics of the basic FSO-1 airplane are as follows: A = 2.02
ℎ ̅
2h/b = 0.36 t/c = 0.05
= 0.32 g
Λ/2
= 35⁰
Problem 10.1 A jet aircraft weighing 56,000lbs has a wing area of 900ft2 and its drag equation is C D = 0.016 + 0.04CL2 (in ground effect). It is desired to operate this aircraft. On an existing runway of 30,000ft (ground-run distance) with concrete pavement ( ) at sea level. If the lift-off speed is 1.2 V S and CL max=1.8, compute the thrust required assuming that the jet engine delivers a constant thrust during the take-off run. V W=0.8 & CL in ground roll = 1.0 .
=0. 0 2
∅=0.
(hint: use equation 10.16 to find Fm, and then use the expression for K to find ln(FS/FLOF)).
Given: W = 56,000ft
CL MAX = 1.8
S = 900ft2
VLOF = 1.2 VS
CD = 0.016 + 0.04CL2
VW = 0.8
SG = 3,000ft
= 0.02
Required: T
∅=0
CL = 1.0
Solution: VLOF = 1.2 VS
=1.2 =1.2 ,. .8 VLOF =204.65 ft/s
CD = 0.016 + 0.04CL2 = 0.016 + 0.04(1)2 CD = 0.056
2 = 2 = 2
2 56, 0 00204. 6 5 = 232.1743000 = 12,149.41 = −
−−∅− −− − −∅ = ∅ ∅
2 −−++ − = () 2 2 − − −−− 2 = 2 ( ) − = 2 −−− −. , −.,1−.−.2 .. = 0.0560.0212,0.0.02377 204.65 900
− −. =1.142
1120=1.142 2732.75 1120=1.142 2732.751.142 2732.751.142 1120=1.142 2732. 7 5 1. 1 42 1120= 1. 1 421 ...− − = =14,090.14
Approximation for Nonzero Windspeed
≠0,. ∫∓ ± ∫ ∫∓ ± ∫ 2 2 ∫∓ ± 2 ln{ } ∓ ± 2 ln{ } { ∓ }± l n 2 ln + l n + ( )± 2 ln( +) ln± 2 ln + ± 2 ln 1 + ± 2 1 ± ln 1 + ( ) 2 1 1 ± ln 1 + [ ] ( 2 ) ±
( ) 2 ± ℎ 1 1 ln 1 + ̅± ̅ ̅ ± ( ) ± 2 ± 2 ( ±2 ±2) 2 ±2 ± 2 ± For Total Tale-Off Time
But,
Subst.
Approximate Method II for SG
⁄2
Assuming that the acceleration vary linearly with
̅
√ 2 1 ̅ ∫∓ ± 1̅ ∫∓ ±̅ ∫∓ 1 2̅ ∓ ±̅ ∓ 1 2̅ ∓ ±̅ ∓ 1 2̅ ( )±̅ ± 1 21̅ ( ±2 +2) 2̅1 ( ±2 +) 2̅ ± ± 2 ̅ = √ ̅ √ 2 ̅ √ 2 ⁄ 2 average acceleration may be calculated at .
to
, or at
, then ts
2 +2 2 ̅ +2 ̅ +2 +1 ± 2 2 ± 2 0 + 0 ̅ √ 2 If
Where
≅3
Approximation Method For Rotation Distance,
Where aircraft).
seconds for modern swept-wing aircraft(less for small
+ 1+ 1
Approximate Method for Trasition Distance, If is the lift coefficient during the transition, then:
where
is the frequently assumed to be 0.8
With R calculated, it is necessary to find the climb angle to determine the trasition distance rate of climb is given by:
.. sin sin sin−1 | | ___2 +___ 0.8 sin It follows that the transition distance,
, is given by:
Approximate Method for Climb Distance,
From Fig. 10.13, it is seen that the climb distance is given by:
50ℎ tan 50ℎ tan ℎℎ 2cos (1cos ) ℎ sin (1cos) ℎ >50 35 0. Where:
If
, then
Landing Analysis
ℎ 35′
′ 5 3 =
ℎ
Fig 10.18
Type equat i o n here.
Flight Path Geometry for Landing Analysis
In most repects the calculation of landing performance is similar to the take-off calculation, varying only in the treatment of the approach and flare and in the consideration of auxiliary stopping devices such as speed brakes. The term “approach” applies on ly to the air distance consists of two parts.
1. A steady-state glide path where the airplane is in the final landing configuration prior to touchdown, and 2. The flare. After touchdown, there is a short ground run (approximately 2 seconds) during which no brakes are applied. The remaining ground run is with full brakes to bring the aircraft to a complete stop, Fig. 10.18, shows the schematic used for the following analysis.
Air Distance, SA
ℎ 50′
50
2 ′ ′ 2
ℎ
Flight Path Geometry for Landing Flare
1.1.3101.15 ℎ ℎ ℎ 21 +∆ Where:
By conservation of Energy:
. .+ . . S 12 ( )+ℎ S S 2 +ℎ 1 2 ____+____ 12 S t t ≅0 3 S t S a ̅+∅ ℎ S Charge in
Where:
(Dirty Airplane)
Free Roll Distance,
The time for the free roll,
may be taken as
It is found that
Breaking Distance,
Using the acceleration formula, the decoration can be written as:
Where
Since
The breaking distance
can be written as:
∓ ± S ∫ ± S ∫∓ 1 S ∫∓ ± ∫∓ 1 2 ∓ ± ∓
[ ] 2 2 2
+ × − (−)− { − + } ∅ ̅ ∅ 0 × − − From
Substitute
Where:
Range and Endurance Range- The total horizontal distance traveled by the aircraft. Endurance- The time that an aircraft can stay aloft. •
•
Propeller- Driven Airplanes Specific Fuel Consumption (sfc), c – Defined as the amount of fuel measured in lbs per second, consumed or required by an engine for each horsepower or pound of thrust developed (lbs of fuel per second, per ft.lbs/sec of power, or per lbs of thrust) •
Brake Specific Fuel Consumption, BSFC- defined as the amount of fuel per hour used for each horsepower (lbs of fuel per brake horses, hour per brake horsepower)
•
at v= 0
at
v= If
Where:
Note that with the brakes applied. At on concrete may be taken to be 0.4 to 0.6
−
Relation Between C and BSFC
− 1,980,000 ̇ = ̇ × BSFC
Since the fuel burned is equal to the decrease in airplane weight, it follows that
eq. ❶
From,
, substitute eqn ❶
eqn ❷
With these basic relations, it is possible to develop expressions and methods for computing range and endurance
× ln × ln × ln ln × ln 375 × × × For best range,
should be
maximum For the case of parabolic drag Breguet’s Formulas for Range and Endurance Range From eqn 2
. × × × At level flight,
Integrating both sides ζρ , C, and CL/CD are all assumed If , ζρ, C, CL and CD are all assumed to have a constant average values throughout the flight.
∫ ∫
Where: W0 = Initial gross weight W1 = Final aircraft weight (= W 0-Wf) Wf = fuel weight used
375× × × Endurance From eqn. ❶
× × × × × × × () × × × × × × At level flight,
Substitute
, C,
are all assumed to have
constant
Average values throughout the flight,
∫ × ∫ − × × − 2 × × [] Eseconds × 2( √ ) 778 × (√ ) E=
For the best endurance,
778 (√ )
Speed for best endurance
3 Where:
Given W= W0 = Wf= Wfused = Wl= W0 – Wfused s= CD=_____ + _____ CL2 (Clean Airplane) BHP= =
̇
Calculate: (a)Rbest and corresponding flight speed (b)Ebest and corresponding flight speed
should
be minimum in the case of parabolic drag polar equation, this yields:
3 4
Report No. 12 Range and Endurance
Solution
̇ 75 ℎ 375 2 Best Range, Rbest : Where:
Flight Speed for R best , VRBest :
Where:
For Best Endurance, Ebest,:
778 (√ )
43 Where:
Flight Speed for Best Endurance, VEBest :
3 Where:
Example 11.1 A Cargo plane has the following characteristics: Initial gross weight= 35,000 lb BSFC= 0.45 lb/ BHP-hr CD = 0.02 + 0.05 CL2 ζ = 0.87 S= 300 ft2 Cruise Altitude= 28,000 ft
This airplane is to carry 3000lb of supply and airdrop it at a distance of 1500 miles away and return to the origin.
Given: Wc = 30 000 lb
CL =
BSFC = 0.45 lb CD = 0.02 + 0.05 CL2
τρ = 0.87 s = 300
ft2
h = 28 000 ft Wsupply = 3000 lb R = 1500 miles
.0.02
CL = 0.632 CD = 2 CD = 2(0.02) = 0.04
.... 26 317.92 lb =
=
Amount of fuel consumed Wf used =
Required: Total amount of fuel consumed, W f used(total), and corresponding flying time, Etotal
Wc -W
= 30000 – 26317.92 Wf used = 3682.08 lb
Initial Flight Speed
Solution:
V=
The computation will be done in two parts:
At h = 28000 ft
a) To destination
τρ ln = ln f τρ = [ ( )] = [ o ( )] R = 375
= 1+ ℎ = 0.0023771+ −. = 0.000957 slug/ft V= . .
4.26
3
V= 575.04 fps
Flying time For optimum range, a/ CD should be kept at maximum, CL =
τρ ) ( E = 778
. . E = 778 0.000957300 . . √ . √
Total Amount of Fuel Consumed Wf used(total) = 3682.08 + 2861.95 Wf used(total) = 6544.03 lb
E = 3.95 hours Total Flying Time b) Return Trip Wc = 26317.92 -3000
Etotal = 3.95 + 4.49 Etotal = 8.44 hours
Wc = 23317.92 lb
= [ o )] ( . . = . .. = 20445.97 lb
Problem 11.1
Determine the maximum range, maximum endurance (and speeds for best range and endurance at 10000 ft) of the following airplane: S = 200 ft2
Amount of Fuel Consumed Wf used =
Wc -W
= 23317.92 – 20455.97 Wf used = 2861.95 lb
Flying time
τρ ( ) . . E = 778 0.000957300 . . √ . √ . E = 778
E = 4.49 hours
W = 10000 lb Wf = 4000 lb BSFC = 0.52 lb/BHP-hr
Τρ = 0.90
Power required characteristics being (at 10000 lb gross weight) V, mph
THPreq’d
403
1350
350
925
300
600
250
400
200
250
175
215
WI = 10000 - 4000
150
200
WI = 6000 lb
140
205
R(miles) = 375
130
220
125
240
τρ ln τρ E(hours) = 778 ( )
Req’d: Vbest and VRbest Ebest and VEbest
Centripetal Force
1. A plane of 3800 lb gross weight is turning at 175 mph
Solution:
with an angle of bank of 50˚.
1+ ℎ
= = 0.0023771+ −. = 0.001755 slug/ft
4.26
(a)What is the centrifugal force? (b)What is the lift? (c)What would be the radius of turn?
3
C =
Given:
L
V = 175 mph X
.
CL =
THPreq’d D= THPreq’d =
D = lbs V = mph D=
=
where v is in graph
= 256.67 ft/sec
=
. ,
W = 3800 lb
Β = 50˚ Req’d: a) CF b) L c) R
Sol’n: a) tan β =
CF = W tan β CF = (3800)( tan 50˚) CF = 4528.66 lb
os = os˚
b) L =
L= 5911.75 lb
c) tan β = R= β . R= ˚.
Minimum Speed in Turns
1. A cub has a minimum flying speed of 39.3 mph in straight level flight. Assuming unlimited engine powers, what is the
minimum speed in a) a 30˚ banked turn, b) a 50˚ banked turn, & c) a 70˚ banked turn? Given: Vs = 39.3 mph
Req’d: a) Vs @ β = 30˚ b) Vs @ β = 50˚
c) Vs @ β = 70˚
R = 1718.14 ft
Sol’n : 2. An airplane is making a 40˚ banked of 565 ft radius. What should be the airspeed?
Given: β = 40˚
Vs =
a) For Vs @ 30˚ Vs =
R =565 ft
Req’d: v
tan β =
v = gRtan v = 32.174565tan40˚ v2 = gRtan
v = 123.50 ft/sec
. = 42.23 mph √ os˚
b) For Vs @ 50˚ Vs =
Sol’n :
os
. = 49.03 mph √ os˚
c) For Vs @ 70˚ Vs =
. = 67.20 mph √ os˚
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