Aeng 411 Notes Merged (2)

AENG 411 Applied Subsonic 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)

2 3

λ2 +λ+1 ) λ+1

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 = 𝑏 2 /𝑠= the ratio of the square of the wingspan to the total wing area.

Following the expression of force and moment equations for airfoils the wing aerodynamic forces & moment are written as: 1 𝐿 = 𝐶𝐿 𝜌𝑉 2 𝑆 = 𝐶𝐿 𝑞𝑆 2 1 𝐷 = 𝐶𝐷 𝜌𝑉 2 𝑆 = 𝐶𝐷 𝑞𝑆 2 1 𝑀 = 𝐶𝑀 𝜌𝑉 2 𝑆𝑐 = 𝐶𝑀 𝑞𝑆𝑐 2

𝑏

2. A = 𝑐 = the ratio of the wingspan to the average chord 𝑠 3. A = 2 = the ratio of the total wing 𝑐

area to the square of the average chord

TAPER RATIO (λ) – defined as the ratio of the tip chord (ct) to the root chord (cr) λ = 𝐶𝑡 /𝐶𝑟

Rectangular Wing 𝑐=

𝐶𝑟 + 𝐶𝑡 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 𝛬𝐿𝐸 𝑜𝑟 𝛬𝑐⁄4 .

A straight, tapered wing of 30ft span has a leading-edge & trailing-edge sweep angle of 45𝑜 & 15𝑜 , respectively. Its total area is 285.3 sq. ft. Find the magnitudes of root chord, tip chord & the mean aerodynamic chord.

𝜆=

𝐶𝑡 4.02′ = = 0.268 𝐶𝑟 15′

𝑐=

2 𝜆2 + 𝜆 + 1 𝐶𝑟 ( ) 3 𝜆+1

45𝑜

𝑐= 𝐶𝑟

15′

𝑐 = 10.57′

15′ 15𝑜

𝑏 = 15′ 2

2 0.2682 +0.268+1 (15′ )[ ] 3 0.268+1

𝐶𝑡

CIRCULATION, DOWNWASH, LIFT AND INDUCED DRAG Induced Drag – part of the drag caused by lift

Sol’n For ΔCAB 𝑏⁄ 2 𝐶𝑟

Tan 45𝑜 = 𝐶𝑟 =

𝑏⁄ 2 tan 45𝑜

=

15 tan 45𝑜

𝐶𝑟 = 15′ For ΔCBD: tan 15𝑜 = 𝐶𝑡 =

𝐶𝑡 𝑏⁄ 2

𝑏 tan 15𝑜 2

𝐶𝑡 = 4.02′

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 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.

𝐷𝑖 = 𝜌𝑠′

𝜔12 2

eq. 5

Since, 𝐷𝑖 = 𝐿𝛼𝑖 2

𝐿=

𝐷𝑖 𝛼𝑖

=

𝐿 = 𝜌𝑠 ′

𝜔 𝜌𝑠′ 1 𝜔 𝑣

2

𝑉𝜔12 2𝜔

eq. 6

Equating for L: 𝜌𝜔1 𝑠 ′ 𝑣 = 𝜌𝑠 ′

(Leave space for figure)

1=

𝑉𝜔12 2𝜔

𝜔1 2𝜔

Let: 𝜔 = downwash for behind the wing S’ = swept area (it is a cross-section) of the airstream taken perpendicular to the direction of motion of the wing.

𝜔1 = 2𝜔

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’

From eq. 4 𝐿 = 𝜌𝜔1 𝑠 ′ 𝑣 𝜋𝑏2 )𝑣 4

𝐿 = 𝜌(2𝜔)( 𝐿 = 𝜌𝑣(𝑠 ′ 𝑣)

𝜔1 𝑣

= 𝜌𝜔1 𝑠′𝑣

eq. 4

2. Energy method -the work done on the air mass per unit time equals the kinetic energy increase per unit time. 𝐷𝑖 𝑣 =

𝜔2 𝜌𝑣𝑠′ 21

𝐿=

𝜋𝜌𝑏2 𝜔𝑣 2

But, 1

𝐿 = 𝐶𝐿 2 𝜌𝑣 2 𝑠

Equating for L: 1 2

𝐶𝐿 𝜌𝑣 2 𝑠 =

𝐶𝐷𝑖 =

𝜋𝜌𝑏2 𝜔𝑣 2

Where: e = span efficiency factor or Oswald’s efficiency factor (=0.85-0.95 for the wing alone)(= 1 for elliptical)

𝐶𝐿 𝑣𝑠 = 𝜋𝑏 2 𝜔 𝐶𝐿 = 𝜋

𝐶𝐿2 𝜋𝐴𝑒

𝑏2 𝜔 𝑠 𝑣

Ae = effective aspect ratio

𝐶𝐿 = 𝜋𝐴𝛼𝑖

TOTAL DRAG COEFFICIENT FOR A WING 𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝐷𝑖

For induced angle of attack, 𝛼𝑖 𝛼𝑖 =

CL 𝜋𝐴

𝐶𝐷 = 𝐶𝐷𝑜 +

, where 𝛼𝑖 is in radian

eqn. 1 Where 𝐶𝐷𝑜 is the (lift independent) sum of skin friction & pressure drag.

For induced drag coefficient, 𝐶𝐷𝑖

Glavert has shown that for rectangular wings, more nearly correct formulas are:

𝐷𝑖 = 𝐿𝛼𝑖 𝐶𝐷𝑖 𝑞𝑠 = 𝐶𝐿 𝑞𝑠𝛼𝑖

𝛼𝑖 =

𝐶𝐷𝑖 = 𝐶𝐿 𝛼𝑖 𝐶𝐿 𝐶𝐷𝑖 = 𝐶𝐿 ( ) 𝜋𝐴 𝐶𝐷𝑖 =

𝐶𝐿2 𝜋𝐴

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.

𝐶𝐷𝑖 = A 3 4 5 6 7 8 9

𝐶𝐿 (1 + 𝜏) 𝜋𝐴 𝐶𝐿2 (1 + 𝛿) 𝜋𝐴 τ 0.11 0.14 0.16 0.18 0.20 0.22 0.23

𝐶𝐿 𝜋𝐴𝑒

𝛿 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 𝐶𝐿 = 0.42, determine: (a) induced angle of attack, (b) induced drag coefficient.

For non-elliptic lift distribution, 𝛼𝑖 =

𝐶𝐿2 𝜋𝐴𝑒

Given: 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑤𝑖𝑛𝑔

45𝑚 𝑠 ℎ = 3000𝑚 𝑒 = 0.85

𝑏 = 14𝑚 𝑐 = 2𝑚 𝐶𝐿 = 0.42

𝑉=

Req’d: (a) 𝛼𝑖 (b) 𝐶𝐷𝑖

Req’d: 𝐷𝑖 = ? Sol’n:

Sol’n: (a) 𝛼𝑖 =

𝐶𝐿 (1 + 𝜋𝐴

1 𝐷𝑖 = 𝐶𝐷𝑖 𝜌𝑉 2 𝑆 2

𝜏), 𝛼 𝑖𝑠 𝑖𝑛 𝑟𝑎𝑑

18.24 𝐶𝐿 (1 + 𝜏), 𝛼𝑖 𝑖𝑛 𝑑𝑒𝑔. 𝐴 180⁄𝜋 → 𝜋 𝑏 14 𝐴= = =7 𝑐 2 From the table: 𝜏 = 0.2 𝛿 = 0.064 @𝐴 =7

=

𝛼𝑖 =

𝐶𝐿2 1 2 𝜌𝑉 𝑆 𝜋𝐴𝑒 2

At level fight: (leave space for figure) 𝐿=𝑊 𝑇=𝐷 1 𝐿 = 𝑊 = 𝐶𝐿 𝜌𝑉 2 𝑆 2 𝐶𝐿 =

(18.24)(0.42) 𝛼𝑖 = [ ] (1 + 0.20) 7

𝑊 1⁄ 𝜌𝑉 2 𝑆 2

𝛼𝑖 = 1.31 𝑑𝑒𝑔. Subst. 𝐶𝐿2

(b) 𝐶𝐷𝑖 = 𝜋𝐴 (1 + 𝛿) =[

(0.42)2 𝜋(7)

] (1 + 0.064)

𝑊 (1 )2 ⁄2 𝜌𝑉 2 𝑆 1 2 𝑊2 𝐷𝑖 = 𝜌𝑉 𝑆 𝐷𝑖 = 𝜋𝐴𝑒 2 𝜋𝐴𝑒 1⁄2 𝜌𝑉 2 𝑆

𝐶𝐷𝑖 = 0.0085 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. 𝑊 = 8500 𝑁 𝑏 = 12𝑚

𝐴=

𝐷𝑖 =

𝑏2 𝑠

𝑊2 𝑏2 1 𝜋 𝑠 𝑒 2 𝜌𝑉 2 𝑆

𝑊 2 ( ) 𝑏 𝐷𝑖 = 1 2 𝜋𝑒 2 𝜌𝑉

For wing # 2 𝛼𝑎2 = 𝛼𝑒 + 𝛼𝑖2 ,

𝛼𝑒 = 𝛼𝑎2 − 𝛼𝑖2

Equating , 𝛼𝑒 = 𝛼𝑒 𝛼𝑎1 − 𝛼𝑖1 = 𝛼𝑎2 − 𝛼𝑖2 𝛼𝑎1 = 𝛼𝑎2 + 𝛼𝑖1 − 𝛼𝑖2 𝐶𝐿 𝐶𝐿 𝛼𝑎1 = 𝛼𝑎2 + − ) 𝜋(𝐴𝑒 1 𝜋(𝐴𝑒 )2 𝐶𝐿 1 1 𝛼𝑎1 = 𝛼𝑎2 + [ − ] 𝜋 (𝐴𝑒 )1 (𝐴𝑒 )2

Where: 𝑊 = 𝑠𝑝𝑎𝑛 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑏 𝑎ℎ 4.26 𝜌 = 𝜌𝑜 (1 + ) 𝑇𝑜 𝑘𝑔⁄ = (1.225 ) 𝑚3 4.26 (−0.00651 𝐾⁄𝑚)(3000𝑚) [1 + ] 288𝐾 𝑘𝑔 𝜌 = 0.908 ⁄ 3 𝑚 8500𝑁 2 ( ) 12𝑚 𝐷𝑖 = 2 1 𝑘𝑔 𝜋(0.85) ( ) (0.908 ⁄ 3 ) (45 𝑚⁄ 2 ) 2 𝑚 𝑠 𝐷𝑖 = 204.37𝑁

Since, 𝐶𝐿 = 𝑎𝛼𝑎 = 𝑎(𝛼 − 𝛼𝑜 ) 𝐶𝐿 𝛼𝑎 = 𝑎 𝐶𝐿 𝐶𝐿 𝐶𝐿 1 1 = + [ − ] 𝑎1 𝑎2 𝜋 (𝐴𝑒 )1 (𝐴𝑒 )2 𝑎

Multiply both sides by 𝐶2 𝐿

𝑎2 𝑎2 1 1 =1+ [ − ] (𝐴 ) (𝐴 𝑎1 𝜋 𝑒 1 𝑒 )2 𝑎2 𝑎1 = 𝑎 1 1 1+ 2[ − 𝜋 (𝐴𝑒 )1 (𝐴𝑒 )2 ]

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 are placed at the one effective angle of attack and 𝛼𝑎 − 𝛼𝑖 , their lift coefficient 𝐶𝐿 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑤ℎ𝑒𝑟𝑒 𝛼𝑎 = 𝛼 − 𝛼𝑜 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑎𝑡𝑡𝑎𝑐𝑘 𝑎𝑛𝑑 𝛼𝑜 𝑖𝑠

Where: 𝑎1 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑜𝑓 𝑤𝑖𝑛𝑔 # 1 𝑎2 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑙𝑖𝑓𝑡 𝑐𝑢𝑟𝑣𝑒 𝑜𝑓 𝑤𝑖𝑛𝑔 # 2 Let: 𝑎1 = 𝑎 𝑎2 = 𝑎∞ 𝑎∞ 𝑎= 𝑎 1 1 1 + 𝜋2 [ − ] (𝐴𝑒 ) (𝐴𝑒 )∞ 𝑎∞ 𝑎= 𝑎 1 + 𝜋𝐴∞ 𝑒

the angle of zero lift.

For wing # 1 𝛼𝑎1 = 𝛼𝑒 + 𝛼𝑖1 ,

Where: 𝛼𝑒 = 𝛼𝑎1 − 𝛼𝑖1

𝑎 = slope of lift curve of wing with finite aspect ratio 𝑎∞ = 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 𝐶𝐿 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 𝑒 = 0.85.

Given: 𝑎∞ =

0.09 𝑑𝑒𝑔

𝐴=7 𝛼𝑎 = 9 𝑑𝑒𝑔. Req’d: 𝐶𝐿 Sol’n 𝐶𝐿 = 𝑎𝛼𝑎 𝑎∞ = 𝑎 ∗ 𝛼𝑎 1 + 𝜋𝐴∞ 𝑒 𝑎∞ 𝛼𝑎 = 𝑎 1 + 𝜋𝐴∞ 𝑒 𝑎∞ 𝛼𝑎 = 18.24𝑎∞ 1+ 𝐴𝑒 0.09 ( )(9 𝑑𝑒𝑔) deg = 0.09 (18.24 deg) ( ) deg 1+ (7)(0.85) 𝐶𝐿 = 0.635

EXAMPLE Prob. 4.2 The test results of the NACA 23012 airfoil show the following.

𝜶𝒐 𝑪𝒍 0 0.15 9 1.2 If this airfoil is used to construct an elliptical wing of 𝐴 = 7.0, determine the wing lift curve slope. Given: NACA 23012 𝜶𝒐 deg 0 9 Elliptical wing (𝑒 = 1)

𝑪𝒍 − 0.15 1.2

Req’d 𝑎@𝐴 =7 (leave space for figure) 𝐶𝐿 = 𝑎 𝛼 𝐶𝑙 − 𝐶𝑙 1 ∆𝐶𝑙 1.2 − 0.15 𝑎∞ = = 2 = ∆𝛼 𝛼2 − 𝛼1 9𝑜 − 0𝑜

𝑎∞ = 0.117⁄𝑑𝑒𝑔 Sol’n: 𝑎=

𝑎=

𝑎∞ 𝑎 1 + 𝜋𝐴∞ 𝑒

𝑎∞ 18.24𝑎∞ 1+ 𝐴𝑒

0.117⁄ 𝑑𝑒𝑔 𝑎= (18.24 deg) (0.117⁄𝑑𝑒𝑔) 1+ 7 𝑎 = 0.090⁄𝑑𝑒𝑔

EXAMPLE PROB. 4.3 A rectangular wing model of 40 in by 5 in has the ff. characteristics determined from a wind tunnel test: 𝑒 = 0.87, 𝐶𝐿𝛼 = 𝑎 = 0.09 𝑝𝑒𝑟 deg & 𝛼𝑜 = −3𝑜 , if a full-scale rectangular wing of 42ft by 6ft is constructed with the same airfoil section, what lift will it develop at 𝛼 = 5𝑜 & 120 𝑚𝑝ℎ under standard sea-level conditions? Assume 𝑒 = 0.87 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑢𝑙𝑙 − 𝑠𝑐𝑎𝑙𝑒 𝑤𝑖𝑛𝑔. Given: Model 𝑏𝑀 = 40𝑖𝑛. 𝑐𝑀 = 5𝑖𝑛 𝐴𝑀 = 8 𝑎𝑀 = 0.09⁄𝑑𝑒𝑔 𝑒𝑀 = 0.87 𝛼𝑜 = −3𝑜 Full Scale 𝑏𝐹𝑆 = 42′ 𝑐𝐹𝑆 = 6′ 𝐴𝐹𝑆 = 7 𝛼𝐹𝑆 = 5𝑜 𝑉𝐹𝑆 = 120𝑚𝑝ℎ SSLC 𝑒𝐹𝑆 = 0.87 𝐿𝐹𝑆 = ? Sol’n 1 𝐿𝐹𝑆 = 𝐶𝐿 𝜌𝑉𝐹𝑆 2 𝑆𝐹𝑆 2 𝐶𝐿 = 𝑎𝐹𝑆 (𝛼𝐹𝑆 − 𝛼𝑜 )

𝑎𝑀 1 1 1 + 18.24𝑎𝑀 [ − ] (𝐴𝑒 )𝐹𝑆 (𝐴𝑒 )𝑀 0.09𝑜 = 1 1 1 + 18.24 (0.09⁄𝑑𝑒𝑔) [ − ] (7)(0.87) (8)(0.87) 𝑎𝐹𝑆 = 0.087⁄𝑑𝑒𝑔 𝑎𝐹𝑆 =

𝐶𝐿 = 0.087⁄𝑑𝑒𝑔 [5𝑜 − (−3𝑜 )] 𝐶𝐿 = 0.696 1 2

L= (0.696) ( ) (0.002377

𝑠𝑙𝑢𝑔 ⁄𝑓𝑡 3 ) (120 ∗

22 𝑓𝑡 2 ) (42𝑓𝑡)(6𝑓𝑡) 15 𝑠𝑒𝑐 2

𝐿𝐹𝑆 = 6,457.05 𝑙𝑏

PART I. FUNDAMENTALS OF FLIGHT MECHANICS FOR STEADY SYMMETRICAL FLIGHT

Fig . Definition of angles and velocities in steady symmetrical flight Where: Xb, Yb, Zb – Body axes system (Yb, not shown, is pointing in the paper), with Xb along some airplane reference line. Xs, Ys, Zs – Stability axes system (Ys 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 + Wsinγ - Tcos(α + θT) = 0 CDqS + Wsinγ - Tcos(α + θT) = 0 + ΣFzs = 0 L + Tsin(α + θT) – Wcosγ = 0 CLqS + Tcos(α + θT) - Wcosγ = 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

1

Part 2

𝐶𝑅 𝑞̅𝑠 = [(𝐶𝐿 𝑞̅𝑠)2 + (𝐶𝐷 𝑞̅𝑠)2 ]2 = 𝜔 2

𝐶𝑅 𝑞̅𝑠 = [(𝐶𝐿 + 𝐶𝐷

Unpowered Flight or Glide

2

𝐶𝑅 𝑞̅𝑠 = (𝐶𝐿 2 + 𝐶𝐷

1 2 )(𝑞̅𝑠) ]2

1 2 2 ) (𝑞̅𝑠)

1

𝐶𝑅 = (𝐶𝐿 2 + 𝐶𝐷 2 )2 =

= 𝜔

= 𝜔

𝜔 𝐸𝑞𝑛. 2 𝑞̅𝑠

R Vv = V cos α Vv = RD = V sin α

β XB XS

∑ 𝐹𝑋𝑜 = 0 W sin α

ZS

Note: 𝜕 >0 𝛼 >0 𝛽>0

𝐷 − 𝜔 sin 𝛼 = 0 𝐷 = 𝜔 sin 𝛼 𝐶𝐷 𝑞̅𝑠 = 𝜔 sin 𝛼 𝐸𝑞𝑛. 3

∑ 𝐹𝑍𝑆 = 0

Fig. 3 Airplane in Gliding Flight

In this flight condition, T = 0 ∑ 𝐹𝑉 = 0 𝑅− 𝜔=0 𝑅= 𝜔

𝐿 − 𝜔 cos 𝛼 = 0 𝐿 = 𝜔 cos 𝛼 𝐶𝐿 𝑞̅𝑠 = 𝜔 cos 𝛼 𝐸𝑞𝑛. 4 Introducing (𝛼̅ = −𝛼), therefore: RD = Vsin 𝛼̅ 𝐸𝑞𝑛. 5 𝐶𝐷 𝑞̅𝑠 = 𝜔 sin 𝛼̅ 𝐸𝑞𝑛. 6 𝐶𝐿 𝑞̅𝑠 = 𝜔 cos 𝛼̅ 𝐸𝑞𝑛. 7

But: 𝑅2 = 𝐿2 − 𝐷2 Substitute; 𝑅2 = 𝜔2 𝐿2 + 𝐷2 = 𝜔2 = 𝑅2 1

𝑅 = (𝐿2 + 𝐷2 )2 = 𝜔

Note that the variables here are 𝜔, h (through 𝜌), 𝜕, v & 𝛼̅. By selecting 𝜔 & h, three variable remain of which one can be arbitrarily selected.

Part IV – Horizontal Distance Covered in a Steady Glide

Where: h

h= Initial altitude R= Horizontal distance covered in a steady glide

R From the given figure tan 𝛾̅ =

Minimum Glide Path Angle γmin & Maximum CL/CD

ℎ 𝑅

𝑅=

ℎ ̅ 𝑡𝑎𝑛𝛾

eqn. 1

𝑅=

ℎ ̅𝑀𝑖𝑛 𝑡𝑎𝑛 𝛾

eqn. 2 𝐶𝐷 𝐶𝐿

But, tan 𝛾 = 𝑅=

It is seen that to achieve the lowest possible 𝛾̅ , it is necessary to maximize CL/CD. For a parabolic drag polar CL/CD can be maximize by setting its derivative with CL to zero.

. Then Substitute to eqn. 1

𝐶𝐷

=0

𝛾 𝐶𝐿

ℎ 𝐶 ( 𝐷)

𝐶𝐷 = 𝐶𝐷𝑜 +

𝐶𝐿

𝐶

𝑅 = ℎ (𝐶 𝐿 ) Hence,

𝛾( 𝐶

𝐷

From, tan 𝛾̅ =

𝑀𝑎𝑥

eqn. 4

𝐶2 𝐿 𝜋𝐴𝑒

)

=𝑜

𝛾 𝐶𝐿 𝑢

*𝑑 (𝑣 ) =

𝐶𝐷 𝐶𝐿

𝑣𝑑𝑢−𝑢𝑑𝑣 𝑣2

𝐶2

𝐶 ( 𝐿⁄𝐶 ) 𝐷 𝑚𝑎𝑥

1 𝐶 ( 𝐿⁄𝐶 ) 𝐷 𝑀𝑎𝑥

1

𝐿 )(1)−(𝐶 )( (𝐶𝐷𝑜 + 𝜋𝐴𝑒 𝐿 𝜋𝐴𝑒 (2𝐶𝐿 ))

1

𝛾̅𝑚𝑖𝑛 = tan−1

𝐶𝐿 𝐶𝐷𝑜 +

𝑅𝑚𝑎𝑥 = ℎ (𝐶 𝐿 )

𝐶𝐿2 𝜋𝐴𝑒

Substitute

eqn.3

𝐷

tan 𝛾̅ =

𝐶 𝛾( 𝐿 )

𝐶2

2

𝐿 ) (𝐶𝐷𝑜 + 𝜋𝐴𝑒

eqn. 5

𝐶𝐷𝑜 +

𝐶𝐿2 𝜋𝐴𝑒

−

2𝐶𝐿2 𝜋𝐴𝑒

𝐶𝐷𝑜 −

𝐶𝐿2 𝜋𝐴𝑒

=0

𝐶𝐷𝑜 =

𝐶𝐿2 𝜋𝐴𝑒

𝐶𝐿2 = 𝜋𝐴𝑒 𝐶𝐷𝑜 𝐶𝐿 = √𝜋𝐴𝑒 𝐶𝐷𝑜

=0

=0

Substitute 𝐶𝐿2 to the polar drag equation 𝐶𝐷 = 𝐶𝐷𝑜 +

3𝐶𝐷𝑜 − 𝐶𝐿2

𝐶𝐿2 𝜋𝐴𝑒 𝜋𝐴𝑒 𝐶𝐷𝑜 𝜋𝐴𝑒

𝐷 𝑚𝑎𝑥 𝐶𝐿 (𝐶 ) 𝐷 𝑚𝑎𝑥

𝐶 𝐶𝐷 𝑚𝑎𝑥

( 𝐿)

Polar Drag Equation

=

√𝜋𝐴𝑒 𝐶𝐷𝑜 2𝐶𝐷𝑜

=

1 𝜋𝐴𝑒 𝐶𝐷𝑜 √ 𝐶 2 2 𝐷𝑜 1 𝜋𝐴𝑒 √ 2 𝐶𝐷𝑜

=

𝐶𝐷 = 𝐶𝐷𝑜 +

Glide Path

𝐶𝐷 = 4𝐶𝐷𝑜 𝐶3

For 𝐶𝐿2 , 𝐶𝐿3 2 𝐶𝐷

𝑤 2 1 √( ) ( ) ( 3 𝑠 𝜌 𝐶𝐿 ( ⁄ 2) 𝐶𝐷

)

𝑚𝑎𝑥

It is seen that to achieve the lowest possible RD, it is 𝑤 𝐶3 necessary to minimize & to maximize 𝐿⁄ 2 . For a 𝑠 𝐶𝐷 3 𝐶𝐿 parabolic drag polar, ⁄ 2 can be maximize by setting 𝐶𝐷 its derivative with CL to zero. 3

𝐶 𝛾 ( 2𝐿 ) 𝐶𝐷

𝛾 𝐶𝐿

=0 𝐶3 𝐿

𝛾

(𝐶𝐷𝑜 +

[

𝐶2 𝐿) 𝜋𝐴𝑒

2

] =0 𝛾𝐶𝐿 𝑢 𝑣𝑑𝑢−𝑢𝑑𝑣 *𝑑 (𝑣 ) = 𝑣2 𝐶2

2

𝐶2

1

𝐿 ) (3𝐶 2 )− 𝐶 3 [2(𝐶 + 𝐿 )( (𝐶𝐷𝑜 + 𝜋𝐴𝑒 𝐷𝑜 𝜋𝐴𝑒 𝜋𝐴𝑒)(2𝐶𝐿 )] 𝐿 𝐿 𝐶2

2 2

𝐿 ) ] [(𝐶𝐷𝑜 + 𝜋𝐴𝑒

(𝐶𝐷𝑜 +

2 𝐶𝐿2 1 (3𝐶𝐿2 ) − ( ) (𝐶𝐷𝑜 ) 𝜋𝐴𝑒 𝜋𝐴𝑒

+

𝐶𝐿2 1 ) (3𝐶𝐿2 ) − (𝜋𝐴𝑒) (4𝐶𝐿4 ) 𝜋𝐴𝑒 𝐶𝐿2 4𝐶 2 3 (𝐶𝐷𝑜 + 𝜋𝐴𝑒 ) − 𝜋𝐴𝑒𝐿 = 0 3𝐶 2 4𝐶 2 3𝐶𝐷𝑜 + 𝜋𝐴𝑒𝐿 − 𝜋𝐴𝑒𝐿 = 0

(𝐶𝐷𝑜 +

𝐶𝐿2 𝜋𝐴𝑒 3𝜋𝐴𝑒 𝐶𝐷𝑜 𝜋𝐴𝑒

𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝐷 = 𝐶𝐷𝑜 + 3𝐶𝐷𝑜

Minimum Rate of Descent

𝑅𝐷𝑚𝑖𝑛 =

= 3𝐶𝐷𝑜 3𝜋𝐴𝑒𝐶𝐷𝑜

𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷𝑜

𝐶𝐷 = 2𝐶𝐷𝑜 𝐶

=0

𝜋𝐴𝑒 𝐶𝐿2 =

𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝐷 = 𝐶𝐷𝑜 + 𝐶𝐷𝑜

(𝐶 𝐿 )

𝐶𝐿2 𝜋𝐴𝑒

𝐶𝐿2 ) (4𝐶𝐿4 ) 𝜋𝐴𝑒

=0

=0

𝐷

=

√(3𝜋𝐴𝑒𝐶𝐷𝑜 )3 (4𝐶𝐷𝑜 )2

𝐶𝐿3 2 𝐶𝐷

=

√(3𝜋𝐴𝑒𝐶𝐷𝑜 )3 2 16𝐶𝐷𝑜

𝐶𝐿3 2 𝐶𝐷

=

3 (3𝜋𝐴𝑒𝐶 ) (3𝜋𝐴𝑒)3 𝐶𝐷𝑜 1 𝐷𝑜 √ 4 16 𝐶𝐷𝑜

𝐶𝐿3 2 𝐶𝐷

=

3 3𝜋𝐴𝑒 (𝜋𝐴𝑒)√ 16 𝐶𝐷𝑜

𝐶3

(𝐶𝐿2 ) 𝐷

𝑚𝑎𝑥

=

3 3𝜋𝐴𝑒 𝜋𝐴𝑒√ 𝐶 16 𝐷𝑜

MINIMUM AIRSPEED OR STALLING SPEED

W

2

1

VMIN = VS = √( s ) (P) (Clmax)

Cl 1 πAe ( ) ⃒ max = √ Cd 2 Cdo

equation 17

=

1 2

1

√(0.022)(0.010)

Cl

(Cd) ⃒ max = 33.71 To achieve a low minimum airspeed (stalling w speed), it is seen that s should be low and CLMAX

Rmax = 1500 ft. (33.71) Rmax = 50 565 ft.

should be high. EXAMPLE PROBLEMS

W s

2 ρ

1 Cl

V @ Rmax = √( ) ( ) ( )

1. A glider weighs 800 lbs. and has a wing loading of 12 psf. Its drag equation is: CD = 0.010 + 0.022 CL2. After being launched at 1500 ft. in still air, find:

CL = √πAeCDO 0.010

= √0.022

a) The greatest distance it can cover b) The greatest duration of flight possible over level ground

CL = 0.674 2

In both cases, find the corresponding flight speeds. Ignore the effect of density changes of the atmosphere and use standard sea level conditions.

V @ Rmax = 122.39 ft/s CL3 3 πAE 3πAE √ 2 ⃒max = 16 CD0 cD

Given: W = 800 lbs. W/s = 12 psf

cl3 3 1 3 ⃒max = ( )√ 2 cd 16 0.022 (0.022)(0.010)

CDo = 0.010 πAe =

1 0.022 3

ρ = 0.002377 slug/ft (S.S.L.C.)

cl3 ⃒max = cl2

995.24 2

h = 1500 ft Required:

RMAX = h(CL/CD) MAX

1

RD⃒min=√(12 psf) (0.002377 slug/ft2) (995.24) RD⃒min= 3.19 ft/s

a) RMAX & V @RMAX b) tMAX & V @ RDMIN Solution:

1

V @ Rmax = √(12 psf) (0.002377 slug/ft3) (0.674)

tmax =

h RDmin

1500 ft 3.19 ft/s

=

tmax =470.22 s

CL =√3πAeCd0

(

3(0.010) 0.022

=√

Cl ) ⃒ max = 21.30 Cd c = Cd max

Rmax = h( L )

CL = 1.17

(21.30)(2000 ft)

Rmax = 42600 ft

w s

2 P

1 CL

c

V @ RDmin =√( ) ( ) ( )

=√(12 psf) (

γmin = tan-1 (CdL )

max

2

1

1

= tan-1 21.30

γmin = 2.69°

slug) (1.17)

0.002377

ft3

V @ RDmin = 92.90 ft/s

ρ @ h = 2000 ft ah

ρ =[1 + To]4.26 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 CD = 0.0340 + 0.0162 CL2. Compute also the minimum glide path angle and minimum rate of descent. Given: W=4500 lbs

=[1 +

3.566x10−3 R )(2000 ft) ft 4.26

(

]

519 R

Ρ= 2.24 x 10-3 slug/ft3 CL2 3 πAE 3πAE √ 2 ⃒max = 16 CD0 cD

cd2 3 1 3 ⃒max = ( )√ 2 cl 16 0.0162 (0.0162)(0.034)

S = b x c = 42 ft x 7 ft =294 ft2 cd2 ⃒max = 854.18 cl2

h = 2000 ft CDO = 0.0340 πAe = 1/0.0162 Required: Rmax, γmin, RDmin Solution:

W

2

1

RDmin = √( s ) (ρ) (Cl2 /Co2 ) 4500 lbs

2

1

=√( 294 ft2 ) (2.24 x 10−3 ) (854.18 ) RDmin = 4.0 ft/s

Cl 1 πAe ( ) ⃒ max = √ Cd 2 Cdo

Cl 1 1 ( ) ⃒ max = √ Cd 2 (0.0162)(0.0340)

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 (h1 + h2) h = 2000 ft CD = 0.0340 + 0.0162CL2 CDO = 0.0340 ΠAe = 1/0.0162 Required: RDmin Solution 1

S = 2 b (h1 + h2) =

1 2

(21 ft)(4 ft + 8 ft)

s = 126 ft2 x 2 s = 252 ft2 W

2

1

RDmin = √( s ) (ρ) (Cl2 /Co2 ) 4500 lbs

2

1

=√( 252 ft2 ) (2.24 x 10−3 ) (854.18 ) RDmin = 4.32 ft/s

Altitude, h(ft) Sea level 5,000 50,000

𝑇𝑉 = 𝐷𝑉 − 𝑊𝑉𝑠𝑖𝑛𝛾

Density ratio, σ But,

𝑅𝐶 = 𝑉𝑠𝑖𝑛𝛾 Then,

STEADY POWERED FLIGHT For stationary (steady), symmetrical powered flight the equations of motion are: 𝑇𝑐𝑜𝑠(𝛼 + 𝜃𝑇 ) − 𝑊 − 𝐷𝑠𝑖𝑛𝛾 = 0 where 𝜃𝑇 = Thrust Orientation Angle and γ = flight path angle and 𝑇𝑠𝑖𝑛(𝛼 + 𝜃𝑟 ) + 𝐿 − 𝑊𝑐𝑜𝑠𝛾 = 0 In the case of conventional airplanes it turns out that the component to other forces in Tsin(α+θT) is small when compared to other forces in Tsin(α+θT) + L – Wcosγ

𝑇𝑉 = 𝐷𝑉 + 𝑊 𝑅. 𝐶. 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

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, 𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛𝛾 And 𝐿 = 𝑤𝑐𝑜𝑠𝛾

In the case of level flight, RC = 0 +

It is useful to write 𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛𝛾 in terms of “work” by multiplication by the airspeed V.

← ∑ 𝐹ℎ = 0 𝑇−𝐷 =0

𝑇 = 𝐷 = 𝐶𝐷 𝑞𝑆 ∑ 𝐹𝑣 = 0 𝐿=𝑊=0 𝐿 = 𝐶𝐿 𝑞𝑆 = 𝑊 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 𝐹𝑉 = 𝐷𝑉 + 𝑊𝑅𝐶 in this case can be written as:

𝑤 2 1 𝑣 = √( ) ( ) ( ) 𝑠 𝜌 𝐶𝐿 The drag in level flight is 𝐷 = 𝐶𝐷 𝑞𝑆 𝐷=

𝑃𝑎𝑣 = 𝑃𝑅𝐸𝑄𝐷 𝑇𝑉 = 𝐷𝑉 = 𝑊𝑅𝐶 𝑃𝑎𝑣 = 𝑃𝑅𝐸𝑄𝐷 = 𝐸𝑃 (excess power) 𝑃𝑎𝑣 = 𝑃𝑟𝑒𝑞𝑑 (level flight) Aerodynamic center – lift Center of gravity – weight This level flight speed follow from equation (𝐿 = 𝐶𝐿 𝑞𝑆 = 𝑤) 𝐶𝐿 𝑞𝑆 = 𝑊 1 𝐶𝐿 𝜌𝑣 2 𝑆 = 𝑊 2

𝐶𝑑 𝑞𝑆 𝑊 𝐶𝐿 𝑞𝑆

𝐷=

𝑇𝑉 = 𝐷𝑉

𝑊 𝐿

𝑊 𝐶𝐿 𝐶𝐷

Therefore, the power required can be written as

𝑃𝑅𝑒𝑞𝑑 = 𝐷𝑉 =

𝐶𝐷 𝑊 2 1 𝑊√( ) ( ) ( ) 𝐶𝐿 𝑆 𝜌 𝐶𝐿

𝑊 2 𝐶𝐷2 𝑃𝑅𝑒𝑞𝑑 = 𝑊√( ) ( ) ( 3 ) 𝑆 𝜌 𝐶𝐿 𝑊 2 1 𝑃𝑅𝑒𝑞𝑑 = 𝑊√( ) ( ) ( 3 2 ) 𝑆 𝜌 𝐶𝐿 ⁄𝐶𝐷

Report No. 6 Calculation of drag and power required for a low speed turboprop transport. Given:

Drag and power required for the case of parabolic drag polars If a drag polar can be represented by

Airplane weight

W = 18,400 kgf

Wing loading

𝑊 𝑆

Altitude

h = 0m

Aerodynamic characteristics

Fig 8.17

Explanation of Table

𝐶𝐷 = 𝐶𝐷𝑜 +

= 220 𝑘𝑔𝑓 ⁄𝑚2

𝐶𝐿2 𝜋𝐴𝑒

The drag at some V is given by 1 𝐶𝐿2 1 2 𝐷 = 𝐶𝐷 𝜌𝑉 2 𝑆 = ( 𝐶𝐷𝑜 + ) 𝜌𝑉 𝑆 2 𝜋𝐴𝑒 2 Or 𝐷 = 𝐷𝑜 + 𝐷𝑖 1 1 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 𝐶𝐷𝑖 𝜌𝑉 2 𝑆 2 2 1 𝐶𝐿2 1 2 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 𝑉 𝑆 2 𝜋𝐴𝑒 2 From: 1 𝐿 = 𝐶𝐿 𝜌𝑉 2 𝑆 = 𝑊 2 𝐶𝐿 =

Substitute:

𝑊 1 2 𝜌𝑉 𝑆 2

2

1 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 2

(1

𝑊

2 2 𝜌𝑉 𝑆

𝜋𝐴𝑒

Flight Speed at Minimum Drag

)

1 2 𝑊2 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 𝑆 + 1 2 𝜋𝐴𝑒 2 𝜌𝑉 2 𝑆 Where:

𝜕𝐷 =0 𝜕𝑉

1 2 𝜌𝑉 𝑆 2

Do = Parasite drag Di = Induced drag

1 𝜕 (𝐶𝐷𝑜 2 𝜌𝑉 2 𝑆 + 𝜕𝑊

1 𝑊2 (−2𝑣 −3 ) = 0 2𝐶𝐷𝑜 𝜌𝑉𝑆 + 1 2 𝜋𝐴𝑒 𝜌𝑆 2

The power required in level flight therefore follows that 1 𝑊2 𝑃𝑅𝑒𝑞𝑑 = 𝐷𝑉 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 1 2 𝜋𝐴𝑒 𝜌𝑉𝑆 2 𝐴=

Since,

Where:

𝑏2 𝑆

𝑊2 ) 1 𝜋𝐴𝑒 𝜌𝑉 2 𝑆 2 =0

𝐶𝐿 = 𝜌𝑉𝑆 −

4𝑤 2 =0 𝜋𝐴𝑒𝜌𝑉 3

𝜋𝐴𝑒𝐶𝐷𝑜 𝜌2 𝑉 4 𝑆 2 − 4𝑤 2 =0 𝜋𝐴𝑒𝜌𝑉 3 𝑆 𝜋𝐴𝑒𝐶𝐷𝑐 𝜌2 𝑉 4 𝑆 2 = 4𝑤 2

1 𝑊2 𝑃𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 𝑏2 1 2 𝜋 𝑆 𝑒 2 𝜌𝑉𝑆

𝑊 2 2 2 1 𝑉 =( ) ( ) ( ) 𝑆 𝜌 𝜋𝐴𝑒𝐶𝐷𝑜

(𝑊 ⁄𝑏)2 1 𝑃𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 1 2 𝜋𝑒 2 𝜌𝑉

𝑊 2 1 𝑉2 = ( ) ( ) 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜

𝑊 𝑏

= 𝑤𝑖𝑛𝑔 𝑠𝑝𝑎𝑛 𝑙𝑜𝑎𝑑𝑖𝑛𝑔

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.

4

𝑊 2 1 𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = √( ) ( ) ( ) 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 1 2 𝑊2 𝐷𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 + 1 2 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆

2

1 𝑊 2 1 𝐷𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 +

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 =

𝑊2 1 𝑊 2 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) (

1 )] √𝜋𝐴𝑒𝐶𝐷𝑜

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 4 𝜌2 𝑆 2 [( 𝑆 ) (𝜌) ( )] + 𝑊 2 √𝜋𝐴𝑒𝐶 𝐷𝑜

1⁄2

1 𝑊 2 1 𝜋𝐴𝑒 𝜌𝑆 [( ) ( ) ( 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 )] 𝑊2 + 𝑊2

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 =

𝐶𝐷𝑜 𝐶𝐷𝑜 𝐷𝑀𝑖𝑛 = 𝑊√ + 𝑊√ 𝜋𝐴𝑒 𝜋𝐴𝑒

3 √(𝑊𝑆) (𝜌) √(𝜋𝐴𝑒) 2 𝐶𝐷𝑜

𝐶𝐷𝑜 𝐷𝑀𝑖𝑛 = 2𝑊 √ 𝜋𝐴𝑒

1 2 𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = 2𝑊 2 √( ) ( ) √ 3 (𝜋𝐴𝑒) 𝑊𝑆 𝜌

𝐷𝑀𝑖𝑛 = 𝐷𝑜 + 𝐷𝑜 𝑊 2 𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = 2𝑊√( ) ( ) √ 3 (𝜋𝐴𝑒) 𝑆 𝜌

𝐷𝑀𝑖𝑛 = 2𝐷𝑜

POWER REQUIRED AT MINIMUM DRAG 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔

1 3 𝑊2 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆+ 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔

1 𝑊 2 1 = 𝐶𝐷𝑜 𝜌𝑉𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜

HIGH SPEED MINIUM POWER REQUIRED 𝜕𝑃𝑅𝑒𝑞𝑑 =0 𝜕𝑉

3⁄2

+

1 𝜕 (𝐶𝐷𝑜 2 𝜌𝑉 3 𝑆 +

𝑊2 1 2

1 𝑊 2 1 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) ( )] √𝜋𝐴𝑒𝐶𝐷𝑜

𝜕𝑉

𝑊2 ) 1 𝜋𝐴𝑒 2 𝜌𝑉𝑆 =0

1 𝑊2 (−𝑉 −2 ) = 0 𝐶𝐷𝑜 𝜌𝑆(3𝑉 2 ) + 1 2 𝜋𝐴𝑒 2 𝜌𝑆

3 2𝑊 2 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 = 1 2 𝜋𝐴𝑒 2 𝜌𝑉 2 𝑆 2 4 2

3𝜋𝐴𝑒𝐶𝐷𝑜 𝜌 𝑉 𝑆 = 4𝑊

2

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 =

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 4 𝜌2 𝑆 2 [( 𝑆 ) (𝜌) ( )] + 𝑊 2 √3𝜋𝐴𝑒𝐶 𝐷𝑜

1⁄2

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 )]

2

𝑊 2 2 2 1 𝑉4 = ( ) ( ) ( ) 𝑆 𝜌 3𝜋𝐴𝑒𝐶𝐷𝑜

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 =

𝑊 2 1 𝑉2 = ( ) ( ) ( ) 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜

1 2 𝑊 + 𝑊2 3 1⁄2

𝑊𝑆𝜌 (𝜋𝐴𝑒)3 [ 2 √ 3𝐶 ] 𝐷𝑜

4 2 3𝐶𝐷𝑜 √ 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝑊 2 √ 3 𝑊𝑆𝜌 (𝜋𝐴𝑒)3

𝑊 2 1 𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = √( ) ( ) ( ) 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜

4 𝑊 2 3𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝑊 √( ) ( ) √ 3 𝑆 𝜌 (𝜋𝐴𝑒)3

POWER MINIMUM REQUIRED 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛

1 3 𝑊2 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆+ 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆 3⁄2

1 𝑊 2 1 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 +

𝑊2 1⁄2

1 𝑊 2 1 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) ( )] √3𝜋𝐴𝑒𝐶𝐷𝑜

DRAG AT MINIMUM POWER REQUIRED 1 2 𝑊2 𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆+ 1 2 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆

1 𝑊 2 1 𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 𝑊2

+ 1 𝜋𝐴𝑒 𝜌𝑆 [ 2

𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝑊√

𝑊2

] 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆

𝐶𝐷𝑜 3𝐶𝐷𝑜 + 𝑊√ 3𝜋𝐴𝑒 𝜋𝐴𝑒

Altitude, h(ft) Sea level 5,000 50,000

𝑇𝑉 = 𝐷𝑉 − 𝑊𝑉𝑠𝑖𝑛𝛾

Density ratio, σ But,

𝑅𝐶 = 𝑉𝑠𝑖𝑛𝛾 Then,

STEADY POWERED FLIGHT For stationary (steady), symmetrical powered flight the equations of motion are: 𝑇𝑐𝑜𝑠(𝛼 + 𝜃𝑇 ) − 𝑊 − 𝐷𝑠𝑖𝑛𝛾 = 0 where 𝜃𝑇 = Thrust Orientation Angle and γ = flight path angle and 𝑇𝑠𝑖𝑛(𝛼 + 𝜃𝑟 ) + 𝐿 − 𝑊𝑐𝑜𝑠𝛾 = 0 In the case of conventional airplanes it turns out that the component to other forces in Tsin(α+θT) is small when compared to other forces in Tsin(α+θT) + L – Wcosγ

𝑇𝑉 = 𝐷𝑉 + 𝑊 𝑅. 𝐶. 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

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, 𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛𝛾 And 𝐿 = 𝑤𝑐𝑜𝑠𝛾

In the case of level flight, RC = 0 +

It is useful to write 𝑇 = 𝐷 + 𝑊𝑠𝑖𝑛𝛾 in terms of “work” by multiplication by the airspeed V.

← ∑ 𝐹ℎ = 0 𝑇−𝐷 =0

𝑇 = 𝐷 = 𝐶𝐷 𝑞𝑆 ∑ 𝐹𝑣 = 0 𝐿=𝑊=0 𝐿 = 𝐶𝐿 𝑞𝑆 = 𝑊 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 𝐹𝑉 = 𝐷𝑉 + 𝑊𝑅𝐶 in this case can be written as:

𝑤 2 1 𝑣 = √( ) ( ) ( ) 𝑠 𝜌 𝐶𝐿 The drag in level flight is 𝐷 = 𝐶𝐷 𝑞𝑆 𝐷=

𝑃𝑎𝑣 = 𝑃𝑅𝐸𝑄𝐷 𝑇𝑉 = 𝐷𝑉 = 𝑊𝑅𝐶 𝑃𝑎𝑣 = 𝑃𝑅𝐸𝑄𝐷 = 𝐸𝑃 (excess power) 𝑃𝑎𝑣 = 𝑃𝑟𝑒𝑞𝑑 (level flight) Aerodynamic center – lift Center of gravity – weight This level flight speed follow from equation (𝐿 = 𝐶𝐿 𝑞𝑆 = 𝑤) 𝐶𝐿 𝑞𝑆 = 𝑊 1 𝐶𝐿 𝜌𝑣 2 𝑆 = 𝑊 2

𝐶𝑑 𝑞𝑆 𝑊 𝐶𝐿 𝑞𝑆

𝐷=

𝑇𝑉 = 𝐷𝑉

𝑊 𝐿

𝑊 𝐶𝐿 𝐶𝐷

Therefore, the power required can be written as

𝑃𝑅𝑒𝑞𝑑 = 𝐷𝑉 =

𝐶𝐷 𝑊 2 1 𝑊√( ) ( ) ( ) 𝐶𝐿 𝑆 𝜌 𝐶𝐿

𝑊 2 𝐶𝐷2 𝑃𝑅𝑒𝑞𝑑 = 𝑊√( ) ( ) ( 3 ) 𝑆 𝜌 𝐶𝐿 𝑊 2 1 𝑃𝑅𝑒𝑞𝑑 = 𝑊√( ) ( ) ( 3 2 ) 𝑆 𝜌 𝐶𝐿 ⁄𝐶𝐷

Report No. 6 Calculation of drag and power required for a low speed turboprop transport. Given:

Drag and power required for the case of parabolic drag polars If a drag polar can be represented by

Airplane weight

W = 18,400 kgf

Wing loading

𝑊 𝑆

Altitude

h = 0m

Aerodynamic characteristics

Fig 8.17

Explanation of Table

𝐶𝐷 = 𝐶𝐷𝑜 +

= 220 𝑘𝑔𝑓 ⁄𝑚2

𝐶𝐿2 𝜋𝐴𝑒

The drag at some V is given by 1 𝐶𝐿2 1 2 𝐷 = 𝐶𝐷 𝜌𝑉 2 𝑆 = ( 𝐶𝐷𝑜 + ) 𝜌𝑉 𝑆 2 𝜋𝐴𝑒 2 Or 𝐷 = 𝐷𝑜 + 𝐷𝑖 1 1 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 𝐶𝐷𝑖 𝜌𝑉 2 𝑆 2 2 1 𝐶𝐿2 1 2 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 𝑉 𝑆 2 𝜋𝐴𝑒 2 From: 1 𝐿 = 𝐶𝐿 𝜌𝑉 2 𝑆 = 𝑊 2 𝐶𝐿 =

Substitute:

𝑊 1 2 𝜌𝑉 𝑆 2

2

1 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 + 2

(1

𝑊

2 2 𝜌𝑉 𝑆

𝜋𝐴𝑒

Flight Speed at Minimum Drag

)

1 2 𝑊2 𝐷 = 𝐶𝐷𝑜 𝜌𝑉 𝑆 + 1 2 𝜋𝐴𝑒 2 𝜌𝑉 2 𝑆 Where:

𝜕𝐷 =0 𝜕𝑉

1 2 𝜌𝑉 𝑆 2

Do = Parasite drag Di = Induced drag

1 𝜕 (𝐶𝐷𝑜 2 𝜌𝑉 2 𝑆 + 𝜕𝑊

1 𝑊2 (−2𝑣 −3 ) = 0 2𝐶𝐷𝑜 𝜌𝑉𝑆 + 1 2 𝜋𝐴𝑒 𝜌𝑆 2

The power required in level flight therefore follows that 1 𝑊2 𝑃𝑅𝑒𝑞𝑑 = 𝐷𝑉 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 1 2 𝜋𝐴𝑒 𝜌𝑉𝑆 2 𝐴=

Since,

Where:

𝑏2 𝑆

𝑊2 ) 1 𝜋𝐴𝑒 𝜌𝑉 2 𝑆 2 =0

𝐶𝐿 = 𝜌𝑉𝑆 −

4𝑤 2 =0 𝜋𝐴𝑒𝜌𝑉 3

𝜋𝐴𝑒𝐶𝐷𝑜 𝜌2 𝑉 4 𝑆 2 − 4𝑤 2 =0 𝜋𝐴𝑒𝜌𝑉 3 𝑆 𝜋𝐴𝑒𝐶𝐷𝑐 𝜌2 𝑉 4 𝑆 2 = 4𝑤 2

1 𝑊2 𝑃𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 𝑏2 1 2 𝜋 𝑆 𝑒 2 𝜌𝑉𝑆

𝑊 2 2 2 1 𝑉 =( ) ( ) ( ) 𝑆 𝜌 𝜋𝐴𝑒𝐶𝐷𝑜

(𝑊 ⁄𝑏)2 1 𝑃𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉 3 𝑆 + 1 2 𝜋𝑒 2 𝜌𝑉

𝑊 2 1 𝑉2 = ( ) ( ) 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜

𝑊 𝑏

= 𝑤𝑖𝑛𝑔 𝑠𝑝𝑎𝑛 𝑙𝑜𝑎𝑑𝑖𝑛𝑔

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.

4

𝑊 2 1 𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = √( ) ( ) ( ) 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 1 2 𝑊2 𝐷𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 + 1 2 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆

2

1 𝑊 2 1 𝐷𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 +

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 =

𝑊2 1 𝑊 2 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) (

1 )] √𝜋𝐴𝑒𝐶𝐷𝑜

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 4 𝜌2 𝑆 2 [( 𝑆 ) (𝜌) ( )] + 𝑊 2 √𝜋𝐴𝑒𝐶 𝐷𝑜

1⁄2

1 𝑊 2 1 𝜋𝐴𝑒 𝜌𝑆 [( ) ( ) ( 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜 )] 𝑊2 + 𝑊2

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 =

𝐶𝐷𝑜 𝐶𝐷𝑜 𝐷𝑀𝑖𝑛 = 𝑊√ + 𝑊√ 𝜋𝐴𝑒 𝜋𝐴𝑒

3 √(𝑊𝑆) (𝜌) √(𝜋𝐴𝑒) 2 𝐶𝐷𝑜

𝐶𝐷𝑜 𝐷𝑀𝑖𝑛 = 2𝑊 √ 𝜋𝐴𝑒

1 2 𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = 2𝑊 2 √( ) ( ) √ 3 (𝜋𝐴𝑒) 𝑊𝑆 𝜌

𝐷𝑀𝑖𝑛 = 𝐷𝑜 + 𝐷𝑜 𝑊 2 𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔 = 2𝑊√( ) ( ) √ 3 (𝜋𝐴𝑒) 𝑆 𝜌

𝐷𝑀𝑖𝑛 = 2𝐷𝑜

POWER REQUIRED AT MINIMUM DRAG 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔

1 3 𝑊2 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆+ 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝐷𝑟𝑎𝑔 𝑆

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛𝐷𝑟𝑎𝑔

1 𝑊 2 1 = 𝐶𝐷𝑜 𝜌𝑉𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √𝜋𝐴𝑒𝐶𝐷𝑜

HIGH SPEED MINIUM POWER REQUIRED 𝜕𝑃𝑅𝑒𝑞𝑑 =0 𝜕𝑉

3⁄2

+

1 𝜕 (𝐶𝐷𝑜 2 𝜌𝑉 3 𝑆 +

𝑊2 1 2

1 𝑊 2 1 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) ( )] √𝜋𝐴𝑒𝐶𝐷𝑜

𝜕𝑉

𝑊2 ) 1 𝜋𝐴𝑒 2 𝜌𝑉𝑆 =0

1 𝑊2 (−𝑉 −2 ) = 0 𝐶𝐷𝑜 𝜌𝑆(3𝑉 2 ) + 1 2 𝜋𝐴𝑒 2 𝜌𝑆

3 2𝑊 2 𝐶𝐷𝑜 𝜌𝑉 2 𝑆 = 1 2 𝜋𝐴𝑒 2 𝜌𝑉 2 𝑆 2 4 2

3𝜋𝐴𝑒𝐶𝐷𝑜 𝜌 𝑉 𝑆 = 4𝑊

2

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 =

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 4 𝜌2 𝑆 2 [( 𝑆 ) (𝜌) ( )] + 𝑊 2 √3𝜋𝐴𝑒𝐶 𝐷𝑜

1⁄2

1 𝑊 2 1 𝜋𝐴𝑒𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 )]

2

𝑊 2 2 2 1 𝑉4 = ( ) ( ) ( ) 𝑆 𝜌 3𝜋𝐴𝑒𝐶𝐷𝑜

𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 =

𝑊 2 1 𝑉2 = ( ) ( ) ( ) 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜

1 2 𝑊 + 𝑊2 3 1⁄2

𝑊𝑆𝜌 (𝜋𝐴𝑒)3 [ 2 √ 3𝐶 ] 𝐷𝑜

4 2 3𝐶𝐷𝑜 √ 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝑊 2 √ 3 𝑊𝑆𝜌 (𝜋𝐴𝑒)3

𝑊 2 1 𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = √( ) ( ) ( ) 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜

4 𝑊 2 3𝐶𝐷𝑜 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝑊 √( ) ( ) √ 3 𝑆 𝜌 (𝜋𝐴𝑒)3

POWER MINIMUM REQUIRED 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛

1 3 𝑊2 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆+ 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆 3⁄2

1 𝑊 2 1 𝑃𝑅𝑒𝑞𝑑𝑀𝑖𝑛 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 +

𝑊2 1⁄2

1 𝑊 2 1 𝜋𝐴𝑒 2 𝜌𝑆 [( 𝑆 ) (𝜌) ( )] √3𝜋𝐴𝑒𝐶𝐷𝑜

DRAG AT MINIMUM POWER REQUIRED 1 2 𝑊2 𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆+ 1 2 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆

1 𝑊 2 1 𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝐶𝐷𝑜 𝜌𝑆 [( ) ( ) ( )] 2 𝑆 𝜌 √3𝜋𝐴𝑒𝐶𝐷𝑜 𝑊2

+ 1 𝜋𝐴𝑒 𝜌𝑆 [ 2

𝐷𝑀𝑖𝑛𝑅𝑒𝑞𝑑 = 𝑊√

𝑊2

] 1 2 𝜋𝐴𝑒 2 𝜌𝑉𝑀𝑖𝑛𝑅𝑒𝑞𝑑 𝑆

𝐶𝐷𝑜 3𝐶𝐷𝑜 + 𝑊√ 3𝜋𝐴𝑒 𝜋𝐴𝑒

CLIMB PERFORMANCE AND SPEED Equations of motion

Fig. Flight Path Geometry and Forces for Airplane in Accelerated Climbing Flight

Normal to flight path: + ∑ 𝐹𝑉 = 0 𝐿 + 𝑇 sin(𝛼𝑤 − 𝜃𝑇 ) − 𝑊 cos 𝛾 − 𝐶. 𝐹 = 0 where C.F. is the centrifugal force (=

𝑊

𝑑𝑉

𝑉 𝑑𝑅). 𝑔 𝑑𝑉

In normal (straight line) climbing flight, since 𝑑𝑅 = 0, the centrifugal force is zero. Assuming small climb angles (𝛾 < 15°) and (𝛼𝑤 − 𝜃𝑇 ) = 0. 𝐿 + 𝑇 sin(𝛼𝑤 − 𝜃𝑇 ) − 𝑊 cos 𝛾 − 𝐶. 𝐹. = 0 𝐿−𝑊 =0 𝐿=𝑊 Along the flight path: +

→ ∑ 𝐹𝐻 = 0

𝑇𝑐𝑜𝑠(∝ 𝑊 − 𝜃𝑇 ) − 𝑊 sin 𝛾 − 𝐷 −

𝑊 𝑑𝑉 ∙ =0 𝑔 𝑑𝑡

𝑊 𝑑𝑉 ∙ =0 𝑔 𝑑𝑡 𝑊 𝑑𝑉 𝑊 sin 𝛾 = 𝑇 − 𝐷 − ∙ 𝑔 𝑑𝑡 𝑇 − 𝐷 1 𝑑𝑉 sin 𝛾 = − ∙ 𝑊 𝑔 𝑑𝑡 𝑇 − 𝑊 sin 𝛾 − 𝐷 −

Since the rate of climb, R.C., is equal to the vertical component of the flight speed: 𝑑ℎ = 𝑣 sin 𝛾 𝑑𝑡 𝑅. 𝐶. sin 𝛾 = 𝑣 𝑅. 𝐶. 𝑇 − 𝐷 1 𝑑𝑉 𝑑ℎ = − ∙ ∙ 𝑉 𝑊 𝑔 𝑑ℎ 𝑑𝑡 𝑅. 𝐶. 𝑇 − 𝐷 1 𝑑𝑉 = − ∙ ∙ 𝑅. 𝐶. 𝑉 𝑊 𝑔 𝑑ℎ (𝑇 − 𝐷)𝑉 𝑉 𝑑𝑉 𝑅. 𝐶. = − ∙ ∙ 𝑅. 𝐶. 𝑊 𝑔 𝑑ℎ (𝑇 − 𝐷)𝑉 𝑉 𝑑𝑉 𝑅. 𝐶. (1 + )= 𝑔 𝑑ℎ 𝑊 𝑅. 𝐶. =

(𝑇 − 𝐷)𝑉 𝑊 𝑅. 𝐶. = 𝑉 𝑑𝑉 1+𝑔 𝑑ℎ Where: 𝑉 𝑑𝑉 = Acceleration factor 𝑔 𝑑ℎ When the climb is made at a constant true airspeed, 𝑑𝑣⁄𝑑ℎ = 0. Therefore:

𝑅. 𝐶. =

(𝑇 − 𝐷)𝑉 𝑊

As written in its present form, eqn. 𝑅. 𝐶. =

(𝑇−𝐷)𝑉 𝑊

is applicable to a jet plane. Since a

propeller-driven airplane is related in terms of power; instead of thrust, eqn. 𝑅. 𝐶. = (𝑇−𝐷)𝑉 𝑊

can be written for propeller-driven airplane as: 𝑇𝑉 − 𝐷𝑉 𝑅. 𝐶. = 𝑊 𝑃𝐴𝑉 − 𝑃𝑅𝑒𝑞′𝑑 𝑅. 𝐶. = 𝑊

𝑤 𝑉 2 𝑤 𝑑𝑉 𝐶. 𝐹. = = 𝑉 𝑔 𝑅 𝑔 𝑑𝑅 (Note: v is constant) CLIMB PERFORMANCE AND SPEED OF PROPELLER DRIVEN AIRPLANES POWER REQUIRED Power required for flight at constant speed is given by: 𝑃𝑅𝐸𝑄′𝐷 = 𝐷𝑉 Since for small γ, 𝑇𝑅𝐸𝑄′𝐷 = 𝐷 =

𝐶𝐷 𝑊 𝐶𝐿

And, 𝑊 2 1 𝑉 = √( ) ( ) ( ) 𝑠 𝜌 𝐶𝐿 Therefore, 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 =

1 𝐶𝐷 𝑊 2 1 𝑊√( ) ( ) ( ) 550 𝐶𝐿 𝑠 𝜌 𝐶𝐿

=

𝑊 𝑊 2 𝐶 2 √( ) ( ) ( 𝐷3 ) 550 𝑠 𝜌 𝐶𝐿

=

𝑊 𝑊 2 𝐶𝐷 √( ) ( ) ∙ 3 550 𝑠 𝜌 𝐶 ⁄2 𝐿

At sea level: 𝑇𝑅𝐸𝑄′ 𝐷 = 𝑉𝐶 = √(

𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 𝑜 =

𝐶𝐷 𝑊 𝐶𝐿 𝑜

𝑊𝑜 2 1 )( )( ) 𝑠 𝜌𝑜 𝐶𝐿

𝑊𝑜 𝑊𝑜 2 𝐶𝐷 √( ) ( ) ∙ 3 550 𝑠 𝜌 𝐶 ⁄2 𝐿

where Wo should be interpreted as a reference weight and not as the weight at sea level. Equating

𝐶𝐷 𝐶𝐿

; 𝑇𝑅𝐸𝑄′ 𝐷 𝑇𝑅𝐸𝑄′ 𝐷 𝑜 = 𝑊 𝑊𝑜 𝑊 𝑇𝑅𝐸𝑄′𝐷 = 𝑇𝑅𝐸𝑄′𝐷 𝑜 ( ) 𝑊𝑜

1

Equating √𝐶 : 𝐿

𝑊 𝜌𝑜 𝑉 = 𝑉𝑜 √ ∙ 𝑊𝑜 𝜌 𝑊 1 = 𝑉𝑜 √( ) (𝜌 ) 𝑊𝑜 ⁄𝜌 𝑜

𝑊 1 = 𝑉𝑜 √( ) ( ) 𝑊𝑜 𝜎 Equating

𝐶𝐷

3 : 𝐶𝐿 ⁄2

𝑇𝐻𝑃𝑅𝐸𝑄′𝐷

𝑊 3 1 √ = 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷𝑜 ( ) ( ) 𝑊𝑜 𝜎

REPORT NO.7 POWER REQUIRED AT SEA LEVEL AND ALTITUDES PROBLEM: A single-engine ______________ aircraft has the following characteristics: 𝑊 = 1653 𝑙𝑏 𝑠 = 116.25 𝑓𝑡 2 𝐶𝐿 𝑚𝑎𝑥 = 1.65 𝐶𝐷 = 0.037 + 0.064𝐶𝐿 2 (Clean Airplane) Compute the power required at sea level, 5000ft, 1000ft, 15000ft, and 20000ft altitudes: FOR MINIMUM SPEED: 𝑊 2 1 𝑉𝑚𝑖𝑛 = √( ) ( ) ( ) 𝑠 𝜌𝑜 𝐶𝐿 𝑚𝑎𝑥 1653 𝑙𝑏 2 1 = √( )( )( ) 2 3 116.25 𝑓𝑡 0.002377 𝑠𝑙𝑢𝑔⁄𝑓𝑡 1.65 𝑉𝑚𝑖𝑛 = 85.15 𝑓𝑡⁄𝑠𝑒𝑐 ×

𝐶𝐿 =

𝐶𝐿 =

𝑊 1 2 2 𝜌𝑜 𝑉𝑜 𝑠 5561.83 𝑉𝑜 2

=

15 = 58.06 𝑚𝑝ℎ 22 1653 𝑙𝑏

𝑠𝑙𝑢𝑔 1 22 𝑓𝑡 2 (2) (0,002377 3 ) (𝑉𝑜 × 𝑠𝑒𝑐 ) (116.25𝑓𝑡) 15 𝑓𝑡

(𝑤ℎ𝑒𝑟𝑒: 𝑉𝑜 𝑖𝑠 𝑖𝑛 𝑚𝑝ℎ) 𝑎ℎ 4.26 𝜎 = (1 + ) 𝑇𝑜 Where: 𝜎 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 𝑎 = −0.003566 °𝑅⁄𝑓𝑡 𝑇𝑜 = 519°𝑅 ℎ = 𝑎𝑛𝑦 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 𝑎𝑏𝑜𝑣𝑒 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 𝑢𝑝 𝑡𝑜 𝑡𝑟𝑜𝑝𝑜𝑝𝑎𝑢𝑠𝑒 𝑖𝑛 𝑓𝑡 TABLE 1 Altitude, h (ft) Density Ratio, σ SEA LEVEL 5000 10000 15000

1.000

20000

EXPLANATION OF TABLE 2 Column No. ①, 𝑉𝑜 = ②,

𝐶𝐿

=

③,

𝐶𝐷

=

④,

𝑇𝑅𝐸𝑄′𝐷

=

⑤,

𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 𝑜

=

⑥, ⑦,

𝑉 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷

= =

Given 5561.83 ①2

0.037 + 0.064②2 ③ ②

𝑊

④×① 375 ① √𝜎 ⑤ √𝜎

Power Available The available thrust power is given by: 𝑇𝐻𝑃𝐴𝑉 = 𝐵𝐻𝑃×𝜁𝑃 Where: BHP = shaft brake horsepower 𝜁𝑃 = 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: 𝐵𝐻𝑃 = 𝐵𝐻𝑃𝐷𝐸𝑆 = 116 𝑁 = 𝑁𝐷𝐸𝑆 = 2800 𝑟𝑝𝑚 𝜁𝑃 𝐷𝐸𝑆 = 78% (𝐴𝑠𝑠𝑢𝑚𝑒𝑑) Solution: At level flight, 𝑇𝐻𝑃𝐴𝑉 = 𝑇𝐻𝑃𝑅𝐸𝑄′ 𝐷 = 𝐵𝐻𝑃 ×𝜁𝑃 = (116)(0.78) = 90.48

By interpolation: V0 mph 130 VDES 140

THPREQ’Do hp 76.41 90.48 92.57

𝑉𝐷𝐸𝑆 − 130 90.48 − 76.41 = 140 − 130 92.57 − 76.41 𝑉𝐷𝐸𝑆 = 138.71 𝑚𝑝ℎ

POWER SPEED COEFFICIENT, CS

5

𝐶𝑆 = √

𝜌𝑉 5 𝑁 2 𝐵𝐻𝑃

where: 𝜌 = 0.002377 𝑠𝑙𝑢𝑔⁄𝑓𝑡 3 𝑉 = 𝑉𝐷𝐸𝑆 = __________ 𝑓𝑡⁄𝑠𝑒𝑐 𝑁 = 𝑁𝐷𝐸𝑆 = ________ 𝑟𝑒𝑣⁄𝑠𝑒𝑐 𝑓𝑡 ∙ 𝑙𝑏 𝐵𝐻𝑃 = 𝐵𝐻𝑃𝐷𝐸𝑆 = ___________ 𝑠𝑒𝑐

In fig. 51: @Cs= _________ and ζP=78%, BLADE ANGLE, β=_________ In fig. 52: 𝑉

@β=_________ and Cs= ________, 𝑁𝐷 = 𝐽 Propeller Diameter, D 𝐷(𝐹𝑇) =

𝑉 𝑁𝐽

where: 𝑉 = 𝑉𝐷𝐸𝑆 = __________ 𝑓𝑡⁄𝑠𝑒𝑐 𝑁 = 𝑁𝐷𝐸𝑆 = ________ 𝑟𝑒𝑣⁄𝑠𝑒𝑐 𝐽 = 𝐴𝑑𝑣𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑜 Explanation of Table: ①,

V

=

Given ①

②,

% Design V

=

③,

% Design N

=

Obtained from Fig. 449

𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑝𝑒𝑒𝑑

④,

N

=

③ x DESIGN N

⑤,

BHP

=

③ x DESIGN BHP

⑥,

𝑉

=

𝑁𝐷 𝑉

⑦,

% DESIGN 𝑁𝐷

=

⑧, ⑨, ⑩,

% DESIGN 𝜁𝑃 𝜁𝑃 𝑇𝐻𝑃𝐴𝑉

= = =

① x 88 ④xD ⑥ 𝐷𝐸𝑆𝐼𝐺𝑁

𝑉 𝑁𝐷

Obtained from Fig. 55a ⑧ x DESIGN 𝜁𝑃 ⑤x⑨

⑪,

N

=

④ x altitude factor (fig.55b)

⑫,

BHP at reduced N

=

𝐷𝑒𝑠𝑖𝑔𝑛 𝐵𝐻𝑃 × ⑪

⑬,

BHP

=

⑫ x ALTITUDE FACTOR (fig. 44b)

⑭,

𝑉

=

𝑁𝐷

𝐷𝑒𝑠𝑖𝑔𝑛 𝑁 ①×88 ⑪×D

⑮,

𝑉

% DESIGN 𝑁𝐷

=

⑭÷Design𝑁𝐷

𝑉

⑯,

% DESIGN 𝜁𝑃

=

⑰, ⑱,

𝜁𝑃 𝑇𝐻𝑃𝐴𝑉

= =

Obtained from fig. 55a ⑯ x DESIGN 𝜁𝑃 ⑬x⑰

CLIMB PERFORMANCE AND SPEED Equations of motion

Fig. Flight Path Geometry and Forces for Airplane in Accelerated Climbing Flight

Normal to flight path: + ∑ 𝐹𝑉 = 0 𝐿 + 𝑇 sin(𝛼𝑤 − 𝜃𝑇 ) − 𝑊 cos 𝛾 − 𝐶. 𝐹 = 0 where C.F. is the centrifugal force (=

𝑊

𝑑𝑉

𝑉 𝑑𝑅). 𝑔 𝑑𝑉

In normal (straight line) climbing flight, since 𝑑𝑅 = 0, the centrifugal force is zero. Assuming small climb angles (𝛾 < 15°) and (𝛼𝑤 − 𝜃𝑇 ) = 0. 𝐿 + 𝑇 sin(𝛼𝑤 − 𝜃𝑇 ) − 𝑊 cos 𝛾 − 𝐶. 𝐹. = 0 𝐿−𝑊 =0 𝐿=𝑊 Along the flight path: +

→ ∑ 𝐹𝐻 = 0

𝑇𝑐𝑜𝑠(∝ 𝑊 − 𝜃𝑇 ) − 𝑊 sin 𝛾 − 𝐷 −

𝑊 𝑑𝑉 ∙ =0 𝑔 𝑑𝑡

𝑊 𝑑𝑉 ∙ =0 𝑔 𝑑𝑡 𝑊 𝑑𝑉 𝑊 sin 𝛾 = 𝑇 − 𝐷 − ∙ 𝑔 𝑑𝑡 𝑇 − 𝐷 1 𝑑𝑉 sin 𝛾 = − ∙ 𝑊 𝑔 𝑑𝑡 𝑇 − 𝑊 sin 𝛾 − 𝐷 −

Since the rate of climb, R.C., is equal to the vertical component of the flight speed: 𝑑ℎ = 𝑣 sin 𝛾 𝑑𝑡 𝑅. 𝐶. sin 𝛾 = 𝑣 𝑅. 𝐶. 𝑇 − 𝐷 1 𝑑𝑉 𝑑ℎ = − ∙ ∙ 𝑉 𝑊 𝑔 𝑑ℎ 𝑑𝑡 𝑅. 𝐶. 𝑇 − 𝐷 1 𝑑𝑉 = − ∙ ∙ 𝑅. 𝐶. 𝑉 𝑊 𝑔 𝑑ℎ (𝑇 − 𝐷)𝑉 𝑉 𝑑𝑉 𝑅. 𝐶. = − ∙ ∙ 𝑅. 𝐶. 𝑊 𝑔 𝑑ℎ (𝑇 − 𝐷)𝑉 𝑉 𝑑𝑉 𝑅. 𝐶. (1 + )= 𝑔 𝑑ℎ 𝑊 𝑅. 𝐶. =

(𝑇 − 𝐷)𝑉 𝑊 𝑅. 𝐶. = 𝑉 𝑑𝑉 1+𝑔 𝑑ℎ Where: 𝑉 𝑑𝑉 = Acceleration factor 𝑔 𝑑ℎ When the climb is made at a constant true airspeed, 𝑑𝑣⁄𝑑ℎ = 0. Therefore:

𝑅. 𝐶. =

(𝑇 − 𝐷)𝑉 𝑊

As written in its present form, eqn. 𝑅. 𝐶. =

(𝑇−𝐷)𝑉 𝑊

is applicable to a jet plane. Since a

propeller-driven airplane is related in terms of power; instead of thrust, eqn. 𝑅. 𝐶. = (𝑇−𝐷)𝑉 𝑊

can be written for propeller-driven airplane as: 𝑇𝑉 − 𝐷𝑉 𝑅. 𝐶. = 𝑊 𝑃𝐴𝑉 − 𝑃𝑅𝑒𝑞′𝑑 𝑅. 𝐶. = 𝑊

𝑤 𝑉 2 𝑤 𝑑𝑉 𝐶. 𝐹. = = 𝑉 𝑔 𝑅 𝑔 𝑑𝑅 (Note: v is constant) CLIMB PERFORMANCE AND SPEED OF PROPELLER DRIVEN AIRPLANES POWER REQUIRED Power required for flight at constant speed is given by: 𝑃𝑅𝐸𝑄′𝐷 = 𝐷𝑉 Since for small γ, 𝑇𝑅𝐸𝑄′𝐷 = 𝐷 =

𝐶𝐷 𝑊 𝐶𝐿

And, 𝑊 2 1 𝑉 = √( ) ( ) ( ) 𝑠 𝜌 𝐶𝐿 Therefore, 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 =

1 𝐶𝐷 𝑊 2 1 𝑊√( ) ( ) ( ) 550 𝐶𝐿 𝑠 𝜌 𝐶𝐿

=

𝑊 𝑊 2 𝐶 2 √( ) ( ) ( 𝐷3 ) 550 𝑠 𝜌 𝐶𝐿

=

𝑊 𝑊 2 𝐶𝐷 √( ) ( ) ∙ 3 550 𝑠 𝜌 𝐶 ⁄2 𝐿

At sea level: 𝑇𝑅𝐸𝑄′ 𝐷 =

𝐶𝐷 𝑊 𝐶𝐿 𝑜

𝑊𝑜 2 1 𝑉𝑜 = √( ) ( ) ( ) 𝑠 𝜌𝑜 𝐶𝐿 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 𝑜 =

𝑊𝑜 𝑊𝑜 2 𝐶𝐷 √( ) ( ) ∙ 3 550 𝑠 𝜌 𝐶 ⁄2 𝐿

where Wo should be interpreted as a reference weight and not as the weight at sea level. Equating

𝐶𝐷 𝐶𝐿

; 𝑇𝑅𝐸𝑄′ 𝐷 𝑇𝑅𝐸𝑄′ 𝐷 𝑜 = 𝑊 𝑊𝑜 𝑊 𝑇𝑅𝐸𝑄′𝐷 = 𝑇𝑅𝐸𝑄′𝐷 𝑜 ( ) 𝑊𝑜

1

Equating √𝐶 : 𝐿

𝑊 𝜌𝑜 𝑉 = 𝑉𝑜 √ ∙ 𝑊𝑜 𝜌 𝑊 1 = 𝑉𝑜 √( ) (𝜌 ) 𝑊𝑜 ⁄𝜌 𝑜

𝑊 1 = 𝑉𝑜 √( ) ( ) 𝑊𝑜 𝜎 Equating

𝐶𝐷

3 : 𝐶𝐿 ⁄2

𝑇𝐻𝑃𝑅𝐸𝑄′𝐷

𝑊 3 1 √ = 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷𝑜 ( ) ( ) 𝑊𝑜 𝜎

ACTIVITY NO.7 POWER REQUIRED AT SEA LEVEL AND ALTITUDES PROBLEM: A single-engine ______________ aircraft has the following characteristics: 𝑊 = 1653 𝑙𝑏 𝑆 = 116.25 𝑓𝑡 2 𝐶𝐿 𝑚𝑎𝑥 = 1.65 𝐶𝐷 = 0.037 + 0.064𝐶𝐿 2 (Clean Airplane) Compute the power required at sea level, 5000ft, 1000ft, 15000ft, and 20000ft altitudes: FOR MINIMUM SPEED: 𝑊 2 1 𝑉𝑚𝑖𝑛 = √( ) ( ) ( ) 𝑠 𝜌𝑜 𝐶𝐿 𝑚𝑎𝑥 1653 𝑙𝑏 2 1 = √( )( )( ) 2 3 116.25 𝑓𝑡 0.002377 𝑠𝑙𝑢𝑔⁄𝑓𝑡 1.65 𝑉𝑚𝑖𝑛 = 85.15 𝑓𝑡⁄𝑠𝑒𝑐 ×

𝐶𝐿 =

𝐶𝐿 =

𝑊 1 2 2 𝜌𝑜 𝑉𝑜 𝑠 5561.83 𝑉𝑜 2

=

15 = 58.06 𝑚𝑝ℎ 22 1653 𝑙𝑏

𝑠𝑙𝑢𝑔 1 22 𝑓𝑡 2 (2) (0,002377 3 ) (𝑉𝑜 × 𝑠𝑒𝑐 ) (116.25𝑓𝑡) 15 𝑓𝑡

(𝑤ℎ𝑒𝑟𝑒: 𝑉𝑜 𝑖𝑠 𝑖𝑛 𝑚𝑝ℎ)

Density Ratio from Sea Level up to Tropopause 𝑎ℎ 4.26 𝜎 = (1 + ) 𝑇𝑜 Where: 𝜎 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑟𝑎𝑡𝑖𝑜 𝑎 = −0.003566 °𝑅⁄𝑓𝑡 𝑇𝑜 = 519°𝑅 ℎ = 𝑎𝑛𝑦 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 𝑎𝑏𝑜𝑣𝑒 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 𝑢𝑝 𝑡𝑜 𝑡𝑟𝑜𝑝𝑜𝑝𝑎𝑢𝑠𝑒 𝑖𝑛 𝑓𝑡

TABLE 1 Density Ratio, σ Altitude, h (ft) SEA LEVEL 1.000 5000 10000 15000 20000

EXPLANATION OF TABLE 2 Column No. : ①, 𝑉𝑜 = ②,

𝐶𝐿

=

③,

𝐶𝐷

=

④,

𝑇𝑅𝐸𝑄′𝐷

=

⑤,

𝑇𝐻𝑃𝑅𝐸𝑄′𝐷 𝑜

=

⑥, ⑦,

𝑉 𝑇𝐻𝑃𝑅𝐸𝑄′𝐷

= =

Given 5561.83 ①2

0.037 + 0.064②2 ③ ②

𝑊

④×① 375 ① √𝜎 ⑤ √𝜎

Power Available The available thrust power is given by: 𝑇𝐻𝑃𝐴𝑉 = 𝐵𝐻𝑃×𝜁𝑃 Where: BHP = shaft brake horsepower 𝜁𝑃 = 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: 𝐵𝐻𝑃 = 𝐵𝐻𝑃𝐷𝐸𝑆 = 116 𝑁 = 𝑁𝐷𝐸𝑆 = 2800 𝑟𝑝𝑚 𝜁𝑃 𝐷𝐸𝑆 = 78% (𝐴𝑠𝑠𝑢𝑚𝑒𝑑) Solution: At level flight, 𝑇𝐻𝑃𝐴𝑉 = 𝑇𝐻𝑃𝑅𝐸𝑄′ 𝐷 = 𝐵𝐻𝑃 ×𝜁𝑃 = (116)(0.78) = 90.48

By interpolation: V0

THPREQ’Do

mph 130 VDES 140

hp 76.41 90.48 92.57

𝑉𝐷𝐸𝑆 − 130 90.48 − 76.41 = 140 − 130 92.57 − 76.41 𝑉𝐷𝐸𝑆 = 138.71 𝑚𝑝ℎ POWER SPEED COEFFICIENT, CS

5

𝐶𝑆 = √

𝜌𝑉 5 𝑁 2 𝐵𝐻𝑃

where: 𝜌 = 0.002377 𝑠𝑙𝑢𝑔⁄𝑓𝑡 3 𝑉 = 𝑉𝐷𝐸𝑆 = __________ 𝑓𝑡⁄𝑠𝑒𝑐 𝑁 = 𝑁𝐷𝐸𝑆 = ________ 𝑟𝑒𝑣⁄𝑠𝑒𝑐 𝑓𝑡 ∙ 𝑙𝑏 𝐵𝐻𝑃 = 𝐵𝐻𝑃𝐷𝐸𝑆 = ___________ 𝑠𝑒𝑐

In fig. 51: @Cs= _________ and ζP=78%, BLADE ANGLE, β=_________ In fig. 52: 𝑉

@β=_________ and Cs= ________, 𝑁𝐷 = 𝐽 Propeller Diameter, D 𝐷(𝐹𝑇) =

𝑉 𝑁𝐽

where: 𝑉 = 𝑉𝐷𝐸𝑆 = __________ 𝑓𝑡⁄𝑠𝑒𝑐 𝑁 = 𝑁𝐷𝐸𝑆 = ________ 𝑟𝑒𝑣⁄𝑠𝑒𝑐 𝐽 = 𝐴𝑑𝑣𝑎𝑛𝑐𝑒 𝑅𝑎𝑡𝑖𝑜 Explanation of Table Column No.:

①,

V

=

Given

①

②,

% Design V

=

③,

% Design N

=

Obtained from Fig. 44a

𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑝𝑒𝑒𝑑

④,

N

=

③ x DESIGN N

⑤,

BHP

=

③ x DESIGN BHP

⑥,

𝑉

=

𝑁𝐷 𝑉

① x 88 ④xD ⑥

⑦,

% DESIGN 𝑁𝐷

=

⑧, ⑨, ⑩, ⑪,

% DESIGN 𝜁𝑃 𝜁𝑃 𝑇𝐻𝑃𝐴𝑉 N

= = = =

Obtained from Fig. 55a ⑧ x DESIGN 𝜁𝑃 ⑤x⑨ ④ x altitude factor (fig.55b)

⑫,

BHP at reduced N

=

𝐷𝑒𝑠𝑖𝑔𝑛 𝐵𝐻𝑃 × ⑪

⑬,

BHP

=

⑫ x ALTITUDE FACTOR (fig. 44b)

⑭,

𝑉

=

𝑁𝐷

𝐷𝐸𝑆𝐼𝐺𝑁

𝑉 𝑁𝐷

𝐷𝑒𝑠𝑖𝑔𝑛 𝑁 ①×88 ⑪×D

⑮,

𝑉

% DESIGN 𝑁𝐷

=

⑭÷Design𝑁𝐷

𝑉

⑯,

% DESIGN 𝜁𝑃

=

⑰, ⑱,

𝜁𝑃 𝑇𝐻𝑃𝐴𝑉

= =

Obtained from fig. 55a ⑯ x DESIGN 𝜁𝑃 ⑬x⑰

ALTITUDE, h Ft 5000 10000 15000 20000

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.

ALTITUDE FACTOR Fig. 44b Fig. E5b

ACTIVITY NO. 9

CLIMB PERFORMANCE AND SPEED

Climb Performance and Speed

From the plotted graphs of the power required and available at given altitudes, various performance parameters can be obtained:

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.

Maximum flight speed Maximum rate of climb Maximum climb angle All corresponding speeds 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 𝑅. 𝐶. = 𝑣 sin 𝛾 sin 𝛾 =

𝑅. 𝐶. 𝑣

𝑅. 𝐶. 𝛾 = sin−1 ( ) 𝑣 𝑅. 𝐶. 𝛾𝑚𝑎𝑥 = sin−1 ( ) 𝑣 𝑚𝑎𝑥 In determining the maximum climb angle, care should be taken to make sure that the speed for

h 𝑉

𝑇𝐻𝑃𝐴𝑉

𝑇𝐻𝑃𝑅𝐸𝑄𝐷

𝐸𝐻𝑃 = 𝑇𝐻𝑃𝐴𝑉 − 𝑇𝐻𝑃𝑅𝐸𝑄𝐷

𝑅. 𝐶. = 𝐸𝐻𝑃 𝑥 33,000 𝑊

mph

hp

hp

hp

ft/min

rate of climb, R.C. (fpm)

600.00 500.00 400.00 300.00 200.00 100.00 0.00 0

50

100

150

Flight speed, v (mph)

At h

𝛾𝑀𝐴𝑋 =

ALTITUDE, h

𝑉𝑀𝐴𝑋

MAX R.C.

FT

MPH

FT/MIN

SEA LEVEL 5,000 ...

sin−1 (

𝑅.𝐶. 𝑉

)

𝑀𝐴𝑋

SPEED BEST FOR CLIMB MAX ANGLE, R.C. 𝛾𝑀𝐴𝑋 MPH

DEG

SPEED FOR BEST CLIMB ANGLE MPH

CLIMB PERFORMANCE FOR CONSTANT THPAV 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: 𝐶𝐷 = 𝐶𝐷𝑂 +

𝐶𝐿2 𝜋𝐴𝑒

𝑃𝐴𝑉 −𝑃𝑅𝐸𝑄𝐷 𝑊

𝑊 𝑊 2 𝐶𝐷 ( )( )∗ 3 √ ⁄ 550 𝑆 𝜌 𝐶𝐿 2

𝑇𝐻𝜌𝑅𝐸𝑄𝐷 ∝

𝐶𝐷 3⁄ 2

Therefore, the condition for 3⁄2

𝐶𝐿

𝐶𝐷

to be

maximum. That is the derivative of 3⁄2

𝐶𝐿

𝐶𝐷

With CL should be zero. 3⁄

𝐶 2 𝜕 ( 𝐶𝐿 ) 𝐷 𝜕𝐶𝐿

=0

3⁄ 2

𝐶𝐿

=0

5⁄

1⁄ 3𝐶𝐿 2 𝐶𝐿 2 3 2 2 ( 𝐶𝐷 𝑂 𝐶𝐿 + −2 = 0) ( 1 ) ⁄ 2 2𝜋𝐴𝑒 𝜋𝐴𝑒 𝐶𝐿 2

𝐶𝐿2 𝐶𝐿2 3𝐶𝐷𝑂 + 3 −4 =0 𝜋𝐴𝑒 𝜋𝐴𝑒

𝐶𝐷𝑂

𝐶𝐿2 =0 𝜋𝐴𝑒

1 𝐶𝐿2 = 𝐶𝐷𝑂 = 3 3𝜋𝐴𝑒

𝐶𝐿2 = 3𝜋𝐴𝑒𝐶𝐷 𝑂 𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷 𝑂 𝐶𝐷 = 𝐶𝐷 𝑂 +

3𝜋𝐴𝑒𝐶𝐷 𝑂 𝜋𝐴𝑒

𝐶𝐷 = 4𝐶𝐷 𝑂

𝐶𝐿

maximum R.C. requires

5⁄

2

) Indicates

that maximum R.C. implies minimum PREQD. However, from eqn. 𝑇𝐻𝜌𝑅𝐸𝑄𝐷 =

𝐶𝐿2 (𝐶𝐷 𝑂 + 𝜋𝐴𝑒 )

3𝐶𝐷𝑂 −

For general aviation aircraft, typical values of a range from 0.7 to 0.85, Eqn.(𝑅. 𝐶. =

3⁄ 𝐶𝐿2 3 1⁄ 2𝐶𝐿 (𝐶𝐷 𝑂 + 𝜋𝐴𝑒 ) (2 𝐶𝐿 2 ) − 𝐶𝐿 2 (𝜋𝐴𝑒 )

𝜕( ) 𝐶𝐿2 𝐶𝐷 𝑂 + 𝜋𝐴𝑒 = 0 𝜕𝐶𝐿

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:

𝐶𝐿 =

W = 36,000-lb S = 450-ft2

=

CD = 0.014 + 0.05 CL2 22

Vmax = 602.6-mph * 15 = 883.81 − fps

𝑊 1 2 2 𝜌𝑣 𝑆 36,000𝑙𝑏

1 (0.002377 𝑠𝑙𝑢𝑔⁄𝑓𝑡 3 )(883.81 𝑓𝑡⁄𝑠𝑒𝑐 )2 (450𝑓𝑡 2 ) 2 𝐶𝐿 = 0.086

SEA LEVEL

𝐶𝐷 = 0.014 + 0.05𝐶𝐿2

THρAV = constant

𝐶𝐷 = 0.014 + (0.05)(0.086)2 𝐶𝐷 = 0.0144

Req’d: Max R.C. and Vmax R.C.

1 𝐷 = 𝐶𝐷 𝜌𝑣 2 𝑆 2 Solution: 𝐷=

Thrust HorsePower, THP (hp)

Sea Level

0.0144 𝐷=( ) (36000𝑙𝑏) 0.0860

THPAV = constant

200

𝐶𝐷 𝑊 𝐶𝐿

150

𝐷 = 6027.91𝑙𝑏

100 50 0 0

50

100

150

Airspeed, V (mph)

At Vmax 𝑇𝐻𝜌𝐴𝑉 = 𝑇𝐻𝜌𝑅𝐸𝑄𝐷 Where: D = lb V = fps

𝐷𝑉 = 550

200

250

(6027.91)(883.81) 550 = 9686.41

𝑇𝐻𝜌𝐴𝑉 = 𝑇𝐻𝜌𝑅𝐸𝑄𝐷 =

At minimum power required: 𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷 𝑂 At maximum R.C. 𝑤 2 1 𝑉𝑚𝑎𝑥𝑅.𝐶. = √( ) ( ) ( ) 𝑠 𝜌 𝐶𝐿

= 1083.03ℎ𝑝

𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷 𝑂 𝑅. 𝐶.𝑚𝑎𝑥 = (

𝐶𝐿 = √3 (

1 ) (0.014) 0.05

𝐶𝐿 = 0.917

𝑤

𝐶𝐷 = 0.056

9686.41 − 1083.03 =( ) (33,000) 36,000

CEILINGS AND TIME TO CLIMB Rate of climb varies in straight proportions with altitude

𝐶𝐷 𝑊 𝐶𝐿

0.056 𝐷=( ) (36000) 0.917 𝐷 = 2,198.47𝑙𝑏

𝑇𝐻𝜌𝑅𝐸𝑄𝐷 𝑚𝑖𝑛 =

𝐷𝑉 (2198.47)(270.93) = 550 550

𝑇𝐻𝜌𝑅𝐸𝑄𝐷 𝑚𝑖𝑛 = 1082.97ℎ𝑝

3⁄

𝐶 2 ( 𝐿 ) 𝐶𝐷 𝑚𝑎𝑥

3

(0.917) ⁄2 . = = 15.68 0.056

𝑇𝐻𝜌𝑅𝐸𝑄𝐷 𝑚𝑖𝑛 =

𝑤 𝑤 2 √( ) ( ) 550 𝑠 𝜌

×

1 3⁄ 𝐶𝐿 2

(𝐶 ) 𝐷 𝑚𝑎𝑥

=

) ×33,000

= 7886.42 𝑓𝑡⁄𝑚𝑖𝑛

𝐶𝐷 = 4𝐶𝐷 𝑂 = (4)(0.014)

𝐷=

𝑇𝐻𝜌𝐴𝑉 − 𝑇𝐻𝜌𝑅𝐸𝑄𝐷 𝑚𝑖𝑛

36000 36000 2 1 √( )( )× 550 450 0.002377 15.68

Using ratio and proportions: 𝐻 ℎ = 𝑅. 𝐶.𝑂 𝑅. 𝐶.𝑂 − 𝑅. 𝐶.ℎ 𝐻=

ℎ 𝑅. 𝐶.𝑂 𝑅. 𝐶.𝑂 − 𝑅. 𝐶.ℎ

Using ratio and proportions: 𝐻 𝐻𝑆 = 𝑅. 𝐶.𝑂 𝑅. 𝐶.𝑂 − 100 𝐻𝑆 =

𝐻(𝑅. 𝐶.𝑂 − 100) 𝑅. 𝐶.𝑂

Example: Given: W

=

R.C.O =

588.33fpm

h

10,000ft

=

R.C.H=10000=

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

1653lb

127.97fpm

Required: H and Hs

Solution:

𝐻= 𝐻=

ℎ 𝑅. 𝐶.𝑂 𝑅. 𝐶.𝑂 − 𝑅. 𝐶.ℎ

(10000𝑓𝑡)(588.33 𝑓𝑡⁄𝑚𝑖𝑛) 588.33 𝑓𝑡⁄𝑚𝑖𝑛 − 127.97 𝑓𝑡⁄𝑚𝑖𝑛 𝐻 = 12,779.78𝑓𝑡

𝐻𝑆 = 𝐻𝑠 =

𝐻(𝑅 ∗ 𝐶𝑜 − 100) 𝑅 ∗ 𝐶𝑜

(12,779.78𝑓𝑡)(588.33 𝑓𝑡⁄𝑚𝑖𝑛 − 100)𝑓𝑡 588.33 𝑓𝑡⁄𝑚𝑖𝑛 𝐻𝑆 = 10,607.57𝑓𝑡

Example #2 Given: 𝑊 = 4,000𝑙𝑏 𝐸𝐻𝑃 𝑎𝑡 𝑠𝑒𝑎 𝑙𝑒𝑣𝑒𝑙 𝐸𝐻𝑃@𝑆𝐿 = 60ℎ𝑝 𝐸𝐻𝑃@ℎ=10,000 = 17ℎ𝑝 Required: 𝐻 𝑎𝑛𝑑 𝐻𝑆 Solution: At sea level 𝑅. 𝐶.𝑜 =

𝐸𝐻𝑃 ∗ 33,000 𝑊

𝑅. 𝐶.𝑜 =

60 ∗ 33,000 4000

𝑅. 𝐶.𝑜 = 495𝑓𝑡/𝑚𝑖𝑛 At h = 10,000 ft 𝑅 ∗ 𝐶ℎ =

𝐸𝐻𝑃ℎ ∗ 33,000 𝑊

𝑅 ∗ 𝐶𝑜 =

17 ∗ 33,000 4000

𝑅 ∗ 𝐶𝑜 = 140.25 𝑓𝑡/𝑚𝑖𝑛 𝐻=

ℎ ∗ 𝑅 ∗ 𝐶𝑜 𝑅 ∗ 𝐶𝑜 − 𝑅 ∗ 𝐶ℎ

𝐻=

(10,000)(495) 495 − 140.25

𝐻 = 13,953.49 𝑓𝑡 𝐻𝑆 = 𝐻𝑆 =

𝐻(𝑅 ∗ 𝐶𝑜 − 100) 𝑅 ∗ 𝐶𝑜

(13,953.49)(495 − 100) 495 𝐻𝑆 = 11,134.60 𝑓𝑡

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: 𝑊 = 20𝑘𝑁 𝑅 ∗ 𝐶𝑜 = 350 𝑚/𝑚𝑖𝑛 𝐻 = 4 𝑘𝑚 Required: 𝐻𝑆 Solution: 𝐻𝑆 = 𝐻𝑆 =

𝐻(𝑅 ∗ 𝐶𝑜 − 30.49) 𝑅 ∗ 𝐶𝑜

(4 𝑘𝑚)(350 − 30.49) 350 𝐻𝑆 = 3.65 𝑘𝑚

TIME TO CLIMB

Using ratio & proportion 𝐻 ℎ = 𝑅 ∗ 𝐶𝑜 𝑅 ∗ 𝐶𝑜 − 𝑅 ∗ 𝐶ℎ 𝑅 ∗ 𝐶ℎ = 𝑅 ∗ 𝐶𝑜 − 𝑅 ∗ 𝐶ℎ =

ℎ ∗ 𝑅 ∗ 𝐶𝑜 𝐻

𝐻 ∗ 𝑅 ∗ 𝐶𝑜 − ℎ ∗ 𝑅 ∗ 𝐶𝑜 𝐻

𝑅 ∗ 𝐶ℎ =

𝑅 ∗ 𝐶𝑜 (𝐻 − ℎ) ℎ

𝑑𝑛 𝑅 ∗ 𝐶𝑜 = (𝐻 − ℎ) 𝑑𝑡 𝐻 ℎ

∫ 𝑑𝑡 = 𝑜

𝑡=−

ℎ 𝐻 𝑑ℎ ∫ 𝑅 ∗ 𝐶𝑜 𝑜 (𝐻 − ℎ)

𝐻 [ln(𝐻 − ℎ)]ℎ𝑜 𝑅 ∗ 𝐶𝑜

𝑡=−

𝐻 [𝑙𝑛(𝐻 − ℎ) − 𝑙𝑛(𝐻 − 𝑜)] 𝑅 ∗ 𝐶𝑜

𝑡=− 𝑡=

𝐻 [𝑙𝑛(𝐻 − ℎ) − 𝑙𝑛𝐻] 𝑅 ∗ 𝐶𝑜

𝐻 [𝑙𝑛𝐻 − 𝑙𝑛(𝐻 − ℎ)] 𝑅 ∗ 𝐶𝑜 𝐻 𝐻 𝑙𝑛 𝑅 ∗ 𝐶𝑜 𝐻 − ℎ

𝑡= Where: 𝑡 = 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑐𝑙𝑖𝑚𝑏 𝑡𝑜 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 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: 𝑅 ∗ 𝐶𝑜 = 350 𝑚/𝑚𝑖𝑛 𝐻 = 4.5 𝑘𝑚 ℎ = 2𝑘𝑚 Required: 𝑡ℎ Solution: 𝑡ℎ = 𝑡ℎ = (

𝐻 𝐻 𝑙𝑛 𝑅 ∗ 𝐶𝑜 𝐻 − ℎ

4,500 𝑚 4.5 𝑘𝑚 ) [𝑙𝑛 ( )] 350 𝑚/𝑚𝑖𝑛 4.5 𝑘𝑚 − 2 𝑘𝑚 𝑡ℎ = 7.56 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

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: 𝑊 = 5,200 𝑙𝑏 𝑅 ∗ 𝐶𝑜 = 850 𝑓𝑡/𝑚𝑖𝑛 𝐻 = 18,500 𝑓𝑡 Required: (A)𝑡𝑆𝐿 𝑡𝑜 ℎ=15,000 𝑓𝑡 (B) 𝑡ℎ=15,000 𝑓𝑡 𝑡𝑜 𝐻 (C)

𝑡𝑆𝐿 𝑡𝑜 𝐻

Solution: (A)Time to climb from SL to h = 15,000 ft 𝑡= 𝑡=

𝐻 𝐻 𝑙𝑛 𝑅 ∗ 𝐶𝑜 𝐻 − ℎ

18,500 𝑓𝑡 18,500 𝑓𝑡 [𝑙𝑛 ( )] 350 𝑓𝑡/𝑚𝑖𝑛 18,500 𝑓𝑡 − 15,000 𝑓𝑡 𝑡 = 36.24 𝑚𝑖𝑛𝑠

(B) Time to climb from h = 15,000 ft to absolute ceiling At h = 15,000 ft 𝑅 ∗ 𝐶ℎ = 𝑅 ∗ 𝐶𝑜 − 𝑅 ∗ 𝐶ℎ = 850 𝑓𝑡/ min −

ℎ ∗ 𝑅 ∗ 𝐶𝑜 𝐻

(15,000 𝑓𝑡)(350 𝑓𝑡/𝑚𝑖𝑛) 18,500 𝑓𝑡

𝑅 ∗ 𝐶ℎ = 160.81 𝑓𝑡/𝑚𝑖𝑛 𝑡=

𝑡=

𝐻 𝐻 𝑙𝑛 𝑅 ∗ 𝐶ℎ 𝐻 − ℎ

18,500 𝑓𝑡 18,500 𝑓𝑡 [𝑙𝑛 ( )] 160.81 𝑓𝑡/𝑚𝑖𝑛 18,500 𝑓𝑡 − 3,500 𝑓𝑡 𝑡 = 24.13 𝑚𝑖𝑛𝑠

(C)

Time to climb from SL to H 𝑡 = 36.24 + 24.13 𝑡 = 60.37 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

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 t = Σ③ min 0

346.77

127.97

0

ꝏ

Time to climb to service ceiling 𝑡𝐻𝑆 =

𝐻 𝐻 𝑙𝑛 𝑅 ∗ 𝐶ℎ 𝐻 − 𝐻𝑆

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:

𝑊 = 22,240 𝑁 𝐸𝑃𝑜 = 56 𝑘𝑊 𝐻𝑆 = 3.66 𝑘𝑚 Required: a.) 𝑅. 𝐶.𝑜ℎ b.) 𝑅. 𝐶.𝐻𝑆 c.) 𝑅. 𝐶.𝐻 Solution: 𝑅. 𝐶.𝑜 = 𝑅. 𝐶.𝑜 =

𝐸𝑃 56,000 𝑊 = 𝑊 22, 240 𝑁

2.52𝑚 60𝑠 ∗ = 151.20 𝑚/𝑚𝑖𝑛 𝑠 1𝑚𝑖𝑛

For H, 𝐻 𝐻𝑆 = 𝑅. 𝐶.𝑜 𝑅. 𝐶.𝑜 − 30.49 𝐻=

(3660𝑚)(157.2𝑚/𝑚𝑖𝑛) 𝐻𝑆 𝑅. 𝐶.𝑜 = 𝑅. 𝐶.𝑜 − 30.49 151.20 − 30.49 𝐻 = 4584.09𝑚

a.) 𝑅. 𝐶.ℎ = 𝑅. 𝐶.𝑜 −

ℎ𝑅.𝐶.𝑜 𝐻

151.20𝑚 (2800𝑚)(151.20𝑚/𝑚𝑖𝑛) − 𝑚𝑖𝑛 4584.09𝑚 𝑅. 𝐶.ℎ = 58.85 𝑚/𝑚𝑖𝑛 b.) 𝑅. 𝐶.𝐻𝑆 = 30.49 𝑚/𝑚𝑖𝑛 c.) 𝑅. 𝐶.𝐻 = 0 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: 𝐻𝑆 = 5𝑘𝑚 𝑅. 𝐶.ℎ =

𝑅. 𝐶.𝑜 = 370

𝑚 𝑚𝑖𝑛

Required: a.) 𝑡ℎ = 3 𝑘𝑚 b.) 𝑡𝐻𝑆 Solution: 𝐻 𝐻𝑆 = 𝑅. 𝐶.𝑜 𝑅. 𝐶.𝑜 − 30.49 𝐻=

𝐻𝑆 𝑅. 𝐶.𝑜 𝑅. 𝐶.𝑜 − 30.49

𝐻=

(5)(370) 370 − 30.49

𝐻 = 5.45 𝑘𝑚 a.)

𝑡𝐻 = 𝑡𝐻 = (

𝐻 𝐻 𝑙𝑛 𝑅. 𝐶.𝑜 𝐻 − ℎ

5450 5.45 ) 𝑙𝑛 [ ] 370 5.45 − 3

𝑡𝐻 = 11.78 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 b.)

𝑡𝐻𝑆 = 𝑡𝐻𝑆 = (

𝐻 𝐻 𝑙𝑛 𝑅. 𝐶.𝑜 𝐻 − 𝐻𝑆

5450 5.45 ) 𝑙𝑛 [ ] 370 5.45 − 5

𝑡𝐻 = 37.43 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

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) 𝑉𝐿𝑂𝐹 ≥ 1.1 𝑉𝑆 𝑉𝐶𝐿 ≥ 1.2 𝑉𝑆 Gear up 500fpm

Far Pt 23 (Civil)

FAR Pt 25 (Commercial)

𝑉𝐿𝑂𝐹 ≥ 1.1 𝑉𝑆 𝑉𝐶𝐿 ≥ 1.1 𝑉𝑆 Gear up 300 fpm

𝑉𝐿𝑂𝐹 ≥ 1.1 𝑉𝑆 𝑉𝐶𝐿 ≥ 1.2 𝑉𝑆 Geardown ½ % @ VLOF

@ S.L (AEO) 100 fpm @S.L (OEI) Field Take-off Length distance Definition over 50’

Notes:

@S.L (AEO)

Gear up 3% @ VCL (OEI)

Take-off 115% of takedistance off distance over 50’ with AEO over 35’ or 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

MIL-C-5011A

FAR Pt 23(Civil)

Speeds

𝑉𝐴 ≥ 1.3 𝑉𝑆 𝑉𝑇𝐷 ≥ 1.1 𝑉𝑆 Landing Distance Over 50’

𝑉𝐴 ≥ 1.3 𝑉𝑆 𝑉𝑇𝐷 ≥ 1.15 𝑉𝑆 Landing Distance Over 50’

Field Length Definition

FAR Pt 25(Commercial) 𝑉𝐴 ≥ 1.3 𝑉𝑆 𝑉𝑇𝐷 ≥ 1.15 𝑉𝑆 Landing Distance Over 50’ divided by 0.6

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 SR

SCL

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 (SG), 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)

𝑉𝐺 = 𝑉 ± 𝑉𝑊

Where: VG = Ground speed V = Airspeed VW = Wind Speed If VG = 0 𝑉 = ± 𝑉𝑊 𝑑𝑆𝐺 𝑑𝑡

= 𝑉 ± 𝑉𝑊

𝑑𝑆𝐺 = (𝑉 ± 𝑉𝑊 )𝑑𝑡

(eq.1)

But,

𝑑𝑡 =

𝑑𝑣 𝑎

(eq.2)

Substitute (eq.2) in (eq.1)

𝑑𝑆𝐺 =

(𝑉± 𝑉𝑊 ) 𝑎

𝑑𝑣

(eq.3)

Integrating both sides

𝑆 ∫0 𝐺 𝑑𝑆𝐺

= 𝑉

𝑉𝐿𝑂𝐹 (𝑉± 𝑉𝑊 ) ∫𝑡̅ 𝑎

𝑆𝐺 = ∫𝑡̅𝑉𝐿𝑂𝐹 𝑊

(𝑉± 𝑉𝑊 ) 𝑎

𝑑𝑣

𝑑𝑣

(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.

𝐹𝑚 =

𝑤 𝑎 = 𝑇 − 𝐷 − 𝜇(𝑊 − 𝐿) − 𝑊∅ 𝑔

𝐹𝑚 =

𝑤 𝑎 = 𝑇 − 𝜇𝑊 − (𝐷 − 𝜇𝐿) − 𝑊∅ 𝑔

𝑎=

𝑇 − 𝜇𝑊 − (𝐷 − 𝜇𝐿) − 𝑊∅ 𝑤 𝑔

(𝑒𝑞. 5)

Substitute (eq.5) in (eq.4)

𝑆𝐺 =

𝑉𝐿𝑂𝐹 (𝑉± 𝑉𝑊 ) ∫𝑡̅𝑉 𝑇−𝜇𝑊−(𝐷−𝜇𝐿)−𝑊∅ 𝑊 𝑤 𝑔

𝑑𝑣

𝑆𝐺 =

(𝑉± 𝑉𝑊 ) 𝑉𝐿𝑂𝐹 𝑤 ∫𝑡̅𝑉 𝑔 𝑥 𝑇−𝜇𝑊−(𝐷−𝜇𝐿)−𝑊∅ 𝑑𝑣 𝑊

Coefficient of relations , 𝜇 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

𝑉𝑊1 𝑉𝑊2

1 ℎ1 7

= (ℎ ) 2

-VW = 50% VW1 +VW = 150% VW1 Where: VW1 = wind speed at h1 VW2 = wind speed at h2 h1 = height of MAC from the ground h2 = height of tower (=50ft)

(eq.6)

Effect of speed on thrust

Sea level

TH𝜌AV

Thrust

VLDF Flight speed ,v (mph)

𝑇𝐻𝜌𝐴𝑉 =

𝑇=

𝑇𝑉𝐿𝑂𝐹 375

375𝑇𝐻𝜌𝐴𝑉 𝑉𝐿𝑂𝐹

Where: T = is in lbs. VLOF = is in mph

h=h1

MAC

Part 7 Effect of the ground on CL and CD

Lift coefficient in ground level effect 𝐶𝐿(𝐼𝐺𝐸) = 𝐶𝐿(𝑂𝐺𝐸)

𝐶𝐿(𝐼𝐺𝐸)∝ − 𝐶𝐿(𝐼𝐺𝐸)∝ ∆∝𝐿 𝐶𝐿(𝑂𝐺𝐸)∝

Where: CL(OGE) = lift coefficient in the appropriate configuration out of ground effect =√𝜋𝐴𝑒𝐶𝐷𝑜

𝐶𝐿(𝑂𝐺𝐸)∝ = 𝐶𝐿(𝐼𝐺𝐸)∝ =

2𝜋𝐴 2+√𝐴2 (1+𝑡𝑎𝑛2 Λ𝑐/2)+4 2𝜋𝐴𝐶𝐻 2+√𝐴2

2 𝑒𝑓𝑓 (1+𝑡𝑎𝑛 Λ𝑐/2)+4

𝑝𝑒𝑟 𝑟𝑎𝑑𝑖𝑎𝑛 𝑝𝑒𝑟 𝑟𝑎𝑑𝑖𝑎𝑛

In fig. 10.8 𝐴

2ℎ

𝐴𝐶𝐻

= 𝑓( ) 𝑏

∆∝𝑜 =

𝑡 2.5655 𝑐

[

ℎ 𝑐

(̅)

−

0.1177 ℎ 2 (̅) 𝑐

]

, 𝑑𝑒𝑔. CT/2

CR/2 Λ 𝐶/2 CL

Drag Coefficient in Ground Effect

𝐶𝐷(𝐼𝐺𝐸) = 𝐶𝐷(𝑂𝐺𝐸) + ∆𝐶𝐷𝑖 Where:

𝐶𝐷(𝑂𝐺𝐸) = ______+ ______ 𝐶𝐿 2

(clean airplane)

𝐶𝐿 = 𝐶𝐿(𝑂𝐺𝐸) ∆𝐶𝐷𝑖 = −𝜎′

𝐶𝐿 2 𝜋𝐴𝑒

, 𝑒 = 1.0

In fig 10.9 ℎ

𝜎′ = 𝑓 ( ) 𝑏 For 0.033 < h/b < 0.25 , the ground influence coefficient 𝜎′ can be estimated from:

𝜎′ =

1−1.32 ℎ/𝑏 1.05−7.4 ℎ/𝑏

Part 8 Approximate method I for SG 𝑤 𝑔

𝑎 = 𝐹𝑆 −

(𝐹𝑆 −𝐹𝐿𝑂𝐷 )𝑉 2 𝑉 2 𝐿𝑂𝐹

(eq.7)

Where: (eq.8)

𝐹𝑆 = 𝑇 − 𝜇𝑊 − 𝑊∅

𝑎𝑡 𝑉 = 0

𝐹𝐿𝑂𝐹 = 𝑇 − 𝜇𝑊 − (𝐷 − 𝜇𝐿) − 𝑊∅

𝑎𝑡 𝑉 = 𝑉𝐿𝑂𝐹 (eq.9)

Approximate for zero wind speed Assume VW = 0 , (eq.4) becomes

𝑆𝐺 = 𝑆𝐺 =

𝑤

𝑉

𝐿𝑂𝐹 ∫ 𝑔 0

𝑤

𝑉 𝑑𝑣 𝐹𝑆 −

(𝐹𝑆 −𝐹𝐿𝑂𝐹 )𝑉2 𝑉2 𝐿𝑂𝐹

𝑉

𝑉 𝑑𝑣

𝐿𝑂𝐹 ∫ 𝐹𝑆 𝑉2 𝐿𝑂𝐹 −(𝐹𝑆 −𝐹𝐿𝑂𝐹 )𝑉2 𝑔 0 𝑉2 𝐿𝑂𝐹

𝑆𝐺 =

𝑤 𝑔

𝑉

𝑉2 𝐿𝑂𝐹 ∫0 𝐿𝑂𝐹

𝑉 𝑑𝑣 𝐹𝑆 𝑉2 𝐿𝑂𝐹 −(𝐹𝑆 −𝐹𝐿𝑂𝐹 )𝑉2

Let

𝑢 = 𝐹𝑆 𝑉 2 𝐿𝑂𝐹 − (𝐹𝑆 − 𝐹𝐿𝑂𝐹 )𝑉 2 𝑑𝑢 = −2(𝐹𝑆 − 𝐹𝐿𝑂𝐹 )𝑉𝑑𝑣

𝑉2 𝐿𝑂𝐹

𝑤

𝑉

𝑆𝐺 = − 2𝑔 [(𝐹 −𝐹 𝑆

𝐿𝑂𝐹

𝑉2 𝐿𝑂𝐹

𝑤

𝐿𝑂𝐹

𝑉2 𝐿𝑂𝐹

𝑤

𝑆𝐺 = − 2𝑔 [(𝐹 −𝐹 𝑆

−2(𝐹𝑆 −𝐹𝐿𝑂𝐹 )𝑉𝑑𝑣𝑉

𝑑𝑣 2 2 𝑆 𝑉 𝐿𝑂𝐹 −(𝐹𝑆 −𝐹𝐿𝑂𝐹 )𝑉 𝑉

𝑆𝐺 = − 2𝑔 [(𝐹 −𝐹 𝑆

𝐿𝑂𝐹 ] ∫ 0 ) 𝐹

𝐿𝑂𝐹

[𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 − (𝐹𝑆 − 𝐹𝐿𝑂𝐹 )𝑉2 }]0 𝐿𝑂𝐹 ] ) ] [ 𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 − (𝐹𝑆 − 𝐹𝐿𝑂𝐹 )𝑉𝐿𝑂𝐹 2 } − )

𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 − (𝐹𝑆 − 𝐹𝐿𝑂𝐹 )(0)2 }]

𝑆𝐺 = −

𝑤 2𝑔

[(

𝑉2 𝐿𝑂𝐹 𝐹𝑆 −𝐹𝐿𝑂𝐹

] [ 𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 − 𝐹𝑆 𝑉𝐿𝑂𝐹 2 + 𝐹𝐿𝑂𝐹 𝑉𝐿𝑂𝐹 2 } − )

𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 }] 𝑉2 𝐿𝑂𝐹

𝑤

𝑆𝐺 = 2𝑔 [(𝐹 −𝐹 𝑆

𝐿𝑂𝐹

𝑉2 𝐿𝑂𝐹

𝑤

𝑆𝐺 = 2𝑔 [(𝐹 −𝐹 𝑆

𝐿𝑂𝐹

𝑉2 𝐿𝑂𝐹

𝑤

𝑆𝐺 = 2𝑔 [(𝐹 −𝐹 𝑆

𝑆𝐺 =

𝑆𝐺 =

] [ 𝑙𝑛{𝐹𝑆 𝑉2 𝐿𝑂𝐹 } − 𝑙𝑛{𝐹𝐿𝑂𝐹 𝑉𝐿𝑂𝐹 2 }] )

𝐿𝑂𝐹

)

] 𝑙𝑛 [𝐹

𝑉2 𝐿𝑂𝐹

2𝑔

𝐹𝐿𝑂𝐹 𝐹𝑆 𝐹𝑆 [ 𝐹 ] 𝑆 𝑙𝑛 𝐹𝐿𝑂𝐹

𝐿𝑂𝐹

1−

𝑤 𝑉2 𝐿𝑂𝐹 𝐹𝑚

Where:

𝐹𝑚 = 𝑘𝐹𝑠

𝐿𝑂𝐹 𝑉𝐿𝑂𝐹

𝐹𝑆

] 𝑙𝑛 [𝐹 )

𝑤

2𝑔

𝐹𝑆 𝑉2 𝐿𝑂𝐹

(eq.11)

(eq.10)

]

2

]

𝑘=

𝐹 1− 𝐿𝑂𝐹 𝐹𝑆 (eq.12) 𝐹 𝑙𝑛 𝑆 𝐹𝐿𝑂𝐹

Example 10.1 The wing characteristics of the basic FSO-1 airplane are as follows: A = 2.02

2h/b = 0.36

ℎ

t/c = 0.05

𝑐̅

= 0.32 g

Λ𝑐/2 = 35⁰

Problem 10.1 A jet aircraft weighing 56,000lbs has a wing area of 900ft2 and its drag equation is CD = 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 (𝜇 = 0.02) at sea level. If the lift-off speed is 1.2 VS and CL max=1.8, compute the thrust required assuming that the jet engine delivers a constant thrust during the take-off run. VW=0.8 &∅ = 0. CL in ground roll = 1.0 . (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

𝜇 = 0.02

CL = 1.0

Required: T

Solution: VLOF = 1.2 VS 𝑊

2

= 1.2√( ) ( ) ( 𝑆 𝜌 𝐶

1

𝐿𝑚𝑎𝑥

56,000

= 1.2√(

900

)

2

1

) (0.002377) (1.8)

VLOF =204.65 ft/s

CD = 0.016 + 0.04CL2 = 0.016 + 0.04(1)2 CD = 0.056

𝑤 𝑉2 𝐿𝑂𝐹

𝑆𝐺 = 2𝑔

𝐹𝑚

2

𝐹𝑚

=

𝑤 𝑉 𝐿𝑂𝐹 = 2𝑔 𝑆 𝐺

(56,000)(204.65)2 (2)(32.174)(3000)

𝐹𝑚 =

12,149.41 𝑙𝑏

𝐹𝑚 =

𝐹𝑆 −𝐹𝐿𝑂𝐹 𝐹 𝑙𝑛𝐹 𝑆

𝐿𝑂𝐹

=

𝑇−𝜇𝑊−𝑊∅−[𝑇−𝜇𝑊−(𝐷−𝜇𝐿)−𝑊∅ ] 𝑙𝑛

𝑇−𝜇𝑊−𝑊∅ 𝑇−𝜇𝑊−(𝐷−𝜇𝐿)−𝑊∅

=

𝑙𝑛

1 2

𝑇−𝜇𝑊−𝑇+𝜇𝑊+(𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆 𝑙𝑛

𝑇−𝜇𝑊

1 𝑇−𝜇𝑊−(𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆 2

𝑇−𝜇𝑊 1 2

𝑇−𝜇𝑊−(𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆 𝑇−𝜇𝑊 1 2

𝑇−𝜇𝑊−(𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆

=

= 𝑒

1 2

(𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆 𝐹𝑚

1 (𝐶𝐷 −𝜇𝐶𝐿 ) 𝜌𝑉2 𝐿𝑂𝐹 𝑆 2 𝐹𝑚

𝑇−(0.02)(56,000) 1 2

𝑇−(0.02)(56,000)−[0.056−(0.02)(1)] (0.002377)(204.65)2 (900) 1

{

[0.056−0.02(1)]( )(0.002377)(204.65)2 (900) 2

𝑒

𝑇−1120 𝑇−2732.75

12,149.41

}

= 1.142

𝑇 − 1120 = 1.142 (𝑇 − 2732.75) 𝑇 − 1120 = 1.142 𝑇 − 2732.75(1.142 ) 2732.75(1.142 ) − 1120 = 1.142 𝑇 − 𝑇 2732.75(1.142 ) − 1120 = 𝑇(1.142 − 1) 2732.75(1.142 )−1120 (1.142−1)

𝑇 = 14,090.14 𝑙𝑏

=𝑇

=

Approximation for Nonzero Windspeed 𝐼𝑓 𝑉𝑤 ≠ 0, 𝐸𝑞𝑛. 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑉𝐿𝐷𝐹 𝑡𝐿𝐷𝐹 𝑣 𝑆𝐺 = ∫ 𝑑𝑣 ± 𝑉𝑤 ∫ 𝑑𝑡 ∓𝑉𝑤 𝑎 0 𝑡𝐿𝐷𝐹 𝑤 𝑉𝐿𝐷𝐹 𝑣𝑑𝑣 = ∫ ± 𝑉𝑤 ∫ 𝑑𝑡 (𝐹𝑠 − 𝐹𝐿𝐷𝐹 )𝑉 2 𝑔 ∓𝑉𝑤 0 𝐹𝑠 − 2 𝑉𝐿𝐷𝐹 2 𝑉𝐿𝐷𝐹 𝑤 𝑉𝐿𝐷𝐹 −2(𝐹𝑠 − 𝐹𝐿𝐷𝐹 )𝑣𝑑𝑣 = − ∫ ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) ∓𝑉𝑤 𝐹𝑆 𝑉𝐿𝐷𝐹 − (𝐹𝑆 − 𝐹𝐿𝐷𝐹 )𝑉 2 2 𝑤 𝑉𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 2 [ln{𝐹𝑆 𝑉𝐿𝐷𝐹 = − − (𝐹𝑆 − 𝐹𝐿𝐷𝐹 )𝑉 2 }] ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) ∓𝑉𝑤 2 𝑤 𝑉𝐿𝐷𝐹 2 2 } [ln{𝐹𝑆 𝑉𝐿𝐷𝐹 = − − (𝐹𝑆 − 𝐹𝐿𝐷𝐹 )𝑉𝐿𝐷𝐹 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 2 − ln{𝐹𝑆 𝑉𝐿𝐷𝐹 − (𝐹𝑆 − 𝐹𝐿𝐷𝐹 )(∓𝑉𝑤 )2 }] ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2 𝑤 𝑉𝐿𝐷𝐹 2 2 2 ) =− − 𝐹𝑆 𝑉𝐿𝐷𝐹 + 𝐹𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 [ln(𝐹𝑆 𝑉𝐿𝐷𝐹 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 2 − ln(𝐹𝑆 𝑉𝐿𝐷𝐹 − 𝐹𝑆 𝑉𝑤 2 + 𝐹𝐿𝐷𝐹 𝑉𝑤 2 )] ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2 𝑤 𝑉𝐿𝐷𝐹 2 2 = − 𝐹𝑆 𝑉𝑤 2 + 𝐹𝐿𝐷𝐹 𝑉𝑤 2 ) − ln 𝐹𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 [ln(𝐹𝑆 𝑉𝐿𝐷𝐹 ] ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 2 2 𝑤 𝑉𝐿𝐷𝐹 𝐹𝑆 𝑉𝐿𝐷𝐹 − 𝐹𝑆 𝑉𝑤 2 + 𝐹𝐿𝐷𝐹 𝑉𝑤 2 = [ln ] ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 𝐹𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 2 𝑤 𝑉𝐿𝐷𝐹 𝐹𝑆 𝑉𝑤 2 𝑉𝑤 2 = {ln [ (1 − 2 ) + 2 ]} ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 𝐹𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 2 𝑤 𝑉𝐿𝐷𝐹 𝐹𝑆 𝐹𝐿𝐷𝐹 𝑉𝑤 2 = ± 𝑉𝑤 𝑡𝐿𝐷𝐹 ln [ (1 − )+ 2 ] 2𝑔 𝐹 (1 − 𝐹𝐿𝐷𝐹 ) 𝐹𝐿𝐷𝐹 𝐹𝑆 𝑉𝐿𝐷𝐹 𝑆 𝐹𝑆 2 𝑤 − 𝑉𝑤 2 ) (𝑉𝐿𝐷𝐹 = ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 𝐹𝐿𝐷𝐹 𝑉𝑤 2 (1 − 𝐹 ) (1 − 2 ) 𝑉𝐿𝐷𝐹 𝑆 𝐹𝑆 𝐹𝑆 𝑉𝑤 2 𝑉𝑤 2 ln [ (1 − 2 ) + 2 ] 𝑉𝐿𝐷𝐹 𝑉𝐿𝐷𝐹 ] [ 𝐹𝐿𝐷𝐹 2 𝑤 (𝑉𝐿𝐷𝐹 − 𝑉𝑤 2 ) 𝑆𝐺 = ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 𝐾𝑤𝐹𝑆

2 𝑤 (𝑉𝐿𝐷𝐹 − 𝑉𝑤 2 ) 𝑆𝐺 = ± 𝑉𝑤 𝑡𝐿𝐷𝐹 2𝑔 𝐹𝑚 𝑤 𝑤ℎ𝑒𝑟𝑒 𝐹𝑚 𝑤 = 𝐾𝑤𝐹𝑆 𝐹𝐿𝐷𝐹 𝑉𝑤 2 (1 − 𝐹 ) (1 − 2 ) 𝑉𝐿𝐷𝐹 𝑆 𝐾𝑤 = 𝐹𝑆 𝑉𝑤 2 𝑉𝑤 2 ln [ − 2 )+ 2 ] 𝐹𝐿𝐷𝐹 (1 𝑉𝐿𝐷𝐹 𝑉𝐿𝐷𝐹

For Total Tale-Off Time 𝑡𝐿𝐷𝐹 =

𝑉𝐿𝐷𝐹 ± 𝑉𝑤 𝑎̅

But, 𝑤 𝑎̅ 𝑔 𝐹𝑚 𝑤 𝑎̅ = 𝑤 𝑔 𝐹𝑚 𝑤 =

Subst. 𝑤 𝑉𝐿𝐷𝐹 ± 𝑉𝑤 𝑔 𝐹𝑚 𝑤 2 𝑤 (𝑉𝐿𝐷𝐹 − 𝑉𝑤 2 ) 𝑤 𝑉𝐿𝐷𝐹 ± 𝑉𝑤 𝑆𝐺 = ± 𝑉𝑤 2𝑔 𝐹𝑚 𝑤 𝑔 𝐹𝑚 𝑤 𝑤 2 = − 𝑉𝑤 2 ± 2𝑉𝑤 𝑉𝐿𝐷𝐹 ± 2𝑉𝑤 ) (𝑉𝐿𝐷𝐹 2𝑔𝐹𝑚 𝑤 𝑤 (𝑉 2 ± 2𝑉𝑤 𝑉𝐿𝐷𝐹 ± 𝑉𝑤 ) = 2𝑔𝐹𝑚 𝑤 𝐿𝐷𝐹 𝑤 (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 )2 𝑆𝐺 = 2𝑔𝐹𝑚 𝑤 𝑡𝐿𝐷𝐹 =

Approximate Method II for SG 2 Assuming that the acceleration vary linearly with 𝑉 2 to 𝑉𝐿𝐷𝐹 , then ts 2 𝑉 average acceleration 𝑎̅ may be calculated at 𝑉 2 = 𝐿𝐷𝐹⁄2, or at 𝑉 = 𝑉𝐿𝐷𝐹 ⁄ . √2

1 𝑉𝐿𝐷𝐹 (𝑉 ± 𝑉𝑤 ) 𝑑𝑣 𝑆𝐺 = ∫ 𝑎̅ ∓𝑉𝑤 1 𝑉𝐿𝐷𝐹 𝑉𝑤 𝑉𝐿𝐷𝐹 = ∫ 𝑣 𝑑𝑣 ± ∫ 𝑑𝑣 𝑎̅ ∓𝑉𝑤 𝑎̅ ∓𝑉𝑤 1 2 𝑉𝐿𝐷𝐹 𝑉𝑤 𝑉𝐿𝐷𝐹 [𝑉 ] = ± [𝑉] 2𝑎̅ ∓𝑉𝑤 𝑎̅ ∓𝑉𝑤 1 2 𝑉𝑤 [𝑉𝐿𝐷𝐹 − (∓𝑉𝑤 )2 ] ± [𝑉𝐿𝐷𝐹 − (∓𝑉𝑤 )] = 2𝑎̅ 𝑎̅ 1 𝑉 𝑤 2 = − 𝑉𝑤 2 ) ± (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 ) (𝑉𝐿𝐷𝐹 2𝑎̅ 𝑎̅ 1 2 = − 𝑉𝑤 2 ± 2𝑉𝐿𝐷𝐹 𝑉𝑤 + 2𝑉𝑤 2 ) (𝑉𝐿𝐷𝐹 2𝑎̅ 1 2 = ± 2𝑉𝐿𝐷𝐹 𝑉𝑤 + 𝑉𝑤 2 ) (𝑉𝐿𝐷𝐹 2𝑎̅ 1 (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 )2 𝑆𝐺 = 2𝑎̅ (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 )2 𝑤 𝑆𝐺 = 2𝑔 [ 𝑇 − 𝜇𝑊 − (𝐶 − 𝜇𝐶 )𝑞̅𝑆 − 𝑊𝜑] 𝑉 𝐷 𝐿 𝑎𝑡 𝑉= 𝐿𝐷𝐹 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 )𝑉 2 𝑤 𝑉 𝐹𝑚 𝑤 = 𝑎̅ = 𝐹𝑆 − 𝑎𝑡 𝑉 = 𝐿𝐷𝐹⁄ 2 𝑔 𝑉𝐿𝐷𝐹 √2 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) (𝑉𝐿𝐷𝐹⁄ ) 𝑤 √2 𝑎̅ = 𝐹𝑆 − 2 𝑔 𝑉𝐿𝐷𝐹 2 (𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 𝑉𝐿𝐷𝐹⁄2 = 𝐹𝑆 − 2 𝑉𝐿𝐷𝐹

2

⁄ √2

𝐹𝑆 − 𝐹𝐿𝐷𝐹 ) 2 2𝐹𝑆 − 𝐹𝑆 + 𝐹𝐿𝐷𝐹 = 2 𝑤 𝐹𝑆 + 𝐹𝐿𝐷𝐹 𝑎̅ = 𝑔 2 𝐹𝑆 + 𝐹𝐿𝐷𝐹 𝑎̅ = 𝑤 2 𝑔 1 (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 )2 𝑆𝐺 = 𝐹 +𝐹 2 ( 𝑆 𝑤𝐿𝐷𝐹 ) 2 𝑔 𝑤 (𝑉𝐿𝐷𝐹 ± 𝑉𝑤 )2 𝑆𝐺 = 𝑤 𝑔 2 𝑔 = 𝐹𝑆 − (

If 𝑉𝑤 = 0 2 𝑤 𝑉𝐿𝐷𝐹 𝑆𝐺 = 𝑔 (𝐹𝑆 + 𝐹𝐿𝐷𝐹 ) Where 𝐹𝑆 = 𝑇 − 𝜇𝑊 − 𝑊𝜑 𝑎𝑡 𝑉 = 0 𝐹𝐿𝐷𝐹 = 𝑇 − 𝜇𝑊 − (𝐶𝐷 − 𝜇𝐶𝐿 )𝑞̅𝑆 − 𝑊𝜑 𝑎𝑡 𝑉 =

𝑉𝐿𝐷𝐹 ⁄ √2

Approximation Method For Rotation Distance, 𝑆𝑅 𝑆𝑅 = 𝑉𝐿𝐷𝐹 𝑡𝑅 Where 𝑡𝑅 ≅ 3 seconds for modern swept-wing aircraft(less for small aircraft). Approximate Method for Trasition Distance, 𝑆𝑇𝑅 If 𝐶𝐿𝑇𝑅 is the lift coefficient during the transition, then: 2 2 𝐶𝐿𝑇𝑅 𝑉𝐿𝐷𝐹 𝑤 𝑉𝐿𝐷𝐹 𝐿=𝑤 2 =𝑤+ 𝑔 𝑅 𝑉𝑆 𝐶𝐿𝑚𝑎𝑥 𝑉𝐿𝐷𝐹 2

= 𝑤(

𝑉𝑆

𝐶𝐿𝑇𝑅

) (𝐶

𝐿𝑚𝑎𝑥

) = 𝑤 (1 +

2 𝑉𝐿𝐷𝐹

𝑔𝑅

)

2 𝑉𝐿𝐷𝐹 𝑅= 𝑉𝐿𝐷𝐹 2 𝐶𝐿𝑇𝑅 𝑔( −1 𝑉𝑆 ) ( 𝐶𝐿𝑚𝑎𝑥 )

where

𝐶𝐿𝑇𝑅 𝐶𝐿𝑚𝑎𝑥

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: (𝑇 − 𝐷)𝑉𝐿𝐷𝐹 𝑅. 𝐶. = 𝑉𝐿𝐷𝐹 sin 𝜃𝐶𝐿 = 𝑤 𝑇−𝐷 sin 𝜃𝐶𝐿 = 𝑤 𝑇−𝐷 𝜃𝐶𝐿 = sin−1 | | 𝑤 𝑉 = 𝑉𝐿𝐷𝐹 1 2 𝐷 = 𝐶𝐷𝑇𝑅 𝜌𝑉𝐿𝐷𝐹 𝑆 2 𝐶𝐷𝑇𝑅 = ____ + ____ 𝐶𝐿 2 𝑇𝑅 (𝐶𝑙𝑒𝑎𝑛 𝐴𝑖𝑟𝑝𝑙𝑎𝑛𝑒) 𝐶𝐿𝑇𝑅 = 0.8 𝐶𝐿𝑚𝑎𝑥 It follows that the transition distance, 𝑆𝑇𝑅 , is given by: 𝑆𝑇𝑅 = 𝑅 sin 𝜃𝐶𝐿

Approximate Method for Climb Distance, 𝑆𝐶 From Fig. 10.13, it is seen that the climb distance is given by: 50 − ℎ 𝑇𝑅 𝑆𝐶𝐿 50 − ℎ 𝑇𝑅 = tan 𝜃𝐶𝐿

tan 𝜃𝐶𝐿 = 𝑆𝐶𝐿

Where: ℎ 𝑇𝑅 = 𝑅 − 2 cos 𝜃𝐶𝐿 ℎ 𝑇𝑅 = 𝑅 − (1 − cos 𝜃𝐶𝐿 ) 𝑆𝑇𝑅 ℎ 𝑇𝑅 = (1 − cos 𝜃𝐶𝐿 ) sin 𝜃𝐶𝐿 If ℎ 𝑇𝑅 > 50′ 𝑜𝑟 35′ , then 𝑆𝐶𝐿 = 0.

Landing Analysis ℎ𝑓 = 35′ 𝑉𝐴

ℎ𝑓 = 35′

Type equation here. 𝑉𝑇𝐷

𝛿𝐹

𝑆𝐴

𝑉=𝐷

𝑉𝑇𝐷

𝑆𝐹𝑅

𝑆𝐵

Fig 10.18 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 only 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

𝛿 2

𝐿 ℎ𝑓 = 50′

𝐷

𝛿 𝑇

𝑅′

𝛾 𝜃 50 𝛿

𝑅′

𝛿 2 𝑡𝑜𝑢𝑐ℎ𝑑𝑜𝑤𝑛

Flight Path Geometry for Landing Flare 𝑉𝐴 = 1.3𝑉𝑆(𝐿) 𝑉𝑇𝐷 = (1.10 − 1.15) 𝑉𝑆(𝐿) Where: 𝑉𝐴 = 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑠𝑝𝑒𝑒𝑑 𝑉𝑇𝐷 = 𝑡𝑜𝑢𝑐ℎ𝑑𝑜𝑤𝑛 𝑠𝑝𝑒𝑒𝑑 𝑉𝑆(𝐿) = 𝑠𝑡𝑎𝑙𝑙 𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝐿𝑎𝑛𝑑𝑖𝑛𝑔 𝐶𝑜𝑛𝑓𝑖𝑔𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝜔 2 1 𝑉𝑆(𝐿) = √( ) ( ) ( ) 𝑆 𝜌 𝐶𝐿𝑚𝑎𝑥 𝛿𝑃 𝐶𝐿𝑚𝑎𝑥 𝛿𝑃 = 𝐶𝐿𝑚𝑎𝑥 + ∆𝑓𝐶𝐿𝑚𝑎𝑥 By conservation of Energy:

Charge in (𝐾. 𝐸. + 𝑃. 𝐸. ) = (𝑅𝑒𝑡𝑎𝑟𝑑𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒)S𝐴 1𝜔 (𝑉𝐴 2 − 𝑉𝑇𝐷 2 ) + 𝑊ℎ𝑓 = (𝐷 − 𝑇)S𝐴 2𝑔 𝜔 𝑉𝐴 2 − 𝑉𝑇𝐷 2 S𝐴 = + ℎ𝑓 ) ( 𝑔−𝑇 2𝑦 Where: 1 𝐷 = 𝐷𝐴 = 𝐶𝐷𝐴 𝜌𝑉𝐴 2 𝑆 2 𝐶𝐷𝐴 = ____ + ____𝐶𝐿 2 𝐴 (Dirty Airplane) 𝐶𝐷𝐴 =

𝜔 1 𝜌𝑉 2 𝑆 2 𝐴

Free Roll Distance, S𝐹𝑅 The time for the free roll, t 𝐹𝑅 may be taken as t 𝐹𝑅 ≅ 0 𝑡𝑜 3 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 It is found that S𝐹𝐷 = 𝑉𝑇𝐷 t 𝐹𝑅 Breaking Distance, S𝐵 Using the acceleration formula, the decoration can be written as: 𝑔 a = − [(𝜇𝑊 − 𝑇𝑒 ) − (𝜇𝐶𝐿 − 𝐶𝐷 )𝑞̅𝑆 + 𝑊∅] 𝜔 Where 𝑇𝑒 = 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑡ℎ𝑟𝑢𝑠𝑡 Since 𝐷𝑒 =

𝑑𝑣 −𝑎

The breaking distance S𝐵 can be written as:

∓𝑉𝑊

S𝐵 = ∫ 𝑉𝑇𝐷

𝑉 ± 𝑉𝑊 𝑑𝑣 −𝑎

𝑉𝑇𝐷

𝑉 ± 𝑉𝑊 𝑑𝑣 𝑎 ∓𝑉𝑊

S𝐵 = ∫

1 𝑉𝑇𝐷 𝑉𝑊 𝑉𝑇𝐷 S𝐵 = ∫ 𝑣𝑑𝑣 ± ∫ 𝑑𝑣 𝑎 ∓𝑉𝑊 𝑎 ∓𝑉𝑊 1 2 𝑉𝑇𝐷 𝑉𝑊 𝑉𝑇𝐷 [𝑉 ] [𝑉] = ± 2𝑎 ∓𝑉𝑊 𝑎 ∓𝑉𝑊

= = = =

2 1 + +𝑉 )2 − ( 𝑉𝑊 ) ] 𝑤 [𝑉𝑇𝐷 [(𝑉 𝑇𝐷 2𝑎 − −𝑎 1 + 𝑉𝑤 + 2 2 (𝑉𝑇𝐷 − 𝑉𝑊 ) (𝑉 𝑉 ) 2𝑎 − 𝑎 𝑇𝐷 − 𝑤 1 + 2 2 (𝑉𝑇𝐷 − 𝑉𝑤2 2𝑉𝑇𝐷 𝑉𝑊 + 2𝑉𝑊 ) 2𝑎 − 1 + 2 2 (𝑉𝑇𝐷 2𝑉 𝑉 + 𝑉𝑊 ) 2𝑎 − 𝑇𝐷 𝑊 2 1

+ − ( 𝑉𝑤 )] −

+ 𝑆𝐵 = 2𝑎 (𝑉𝑇𝐷 𝑉𝑤 ) −

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)

From 𝑤 𝐹𝑚𝑤 = 𝑔 𝑎 𝑎=

𝐹𝑚𝑤 𝑤 𝑔

Substitute 𝑆𝐵 =

Range and Endurance • Range- The total horizontal distance traveled by the aircraft. • Endurance- The time that an aircraft can stay aloft.

2 + (𝑉𝑇𝐷 𝑉𝑊 ) − × 𝐹 2𝑔 𝑚𝑤

𝑤

Where: 𝐹𝑚𝑤 = 𝐹𝑠 𝐹𝑚𝑤 = 𝐹𝑠 {

𝐹𝑆 𝑉𝑇𝐷 𝑉2 𝑉2 𝐹𝑠 𝑙𝑛[ (1− 2𝑤 )+ 2𝑤 ] 𝐹𝐵 𝑉𝑇𝐷 𝑉𝑇𝐷

}

𝐹𝑠 = 𝑀𝑊 − 𝑇𝑒 + 𝑊∅ at v= 0 𝐹𝐵 = 𝑀𝑊 − 𝑇𝑒 − (𝑀𝐶𝐿 − 𝐶𝐷 )𝑞̅𝑠 + 𝑊∅ at v= 𝑉𝑇𝐷 If 𝑉𝑤 = 0 𝑤

2 𝑉𝑇𝐷

𝑆𝐵 = 2𝑔 × 𝐹𝑚 Where: 𝐹𝑚 =

𝐹𝑠 −𝐹𝐵 𝑙𝑛

𝐹𝑠 𝐹𝐿𝑂𝐹

𝐹 1− 𝐵 𝐹𝑆 𝐹 ln 𝑆 𝐹𝐵

= 𝐹𝑆 (

)

Note that with the brakes applied. At on concrete may be taken to be 0.4 to 0.6 𝐹𝑚 = 𝐾𝐹𝑠 𝐾=

𝐹 1− 𝐵 𝐹𝑆 𝐹 𝑙𝑛 𝑆 𝐹𝐵

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)

•

2

𝐹 𝑉 (1− 𝐵 )(1− 2𝑤 )

Relation Between C and BSFC 𝑙𝑏 BSFC(𝐵𝐻𝑃−ℎ𝑟) = 𝐶 𝑥 1,980,000

Since the fuel burned is equal to the decrease in airplane weight, it follows that 𝑑𝑤 = −𝐶𝐵𝐻𝑃 𝑑𝑡 𝑑𝑡

𝑑𝑤

1

= − 𝐶𝐵𝐻𝑃

𝐵𝑆𝐹𝐶 = ̇ 𝑊𝑓=

𝑊̇ 𝑓

𝐵𝐻𝑃 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑙𝑏𝑠 ℎ𝑟

× ℎ𝑟

𝑑𝑤

𝑑𝑡 = − 𝐶𝐵𝐻𝑃 eq. ❶ From, 𝑑𝑠 𝑣 = 𝑑𝑡 𝑑𝑡 =

𝑑𝑠 𝑣

, substitute eqn ❶

𝑑𝑠 𝑣

𝑑𝑤

= − 𝐶𝐵𝐻𝑃

𝜁𝜌

𝑅= −

𝑣𝑑𝑤

𝑑𝑠 = − 𝐶𝐵𝐻𝑃 eqn ❷

𝑅=

With these basic relations, it is possible to develop expressions and methods for computing range and endurance

𝑅=

𝑐

𝜁𝜌

0 [ln 𝑊]𝑊 𝑊1

𝐶𝐷 𝐶𝐿

×

𝑐

1 [ln 𝑊]𝑊 𝑊0

𝐶𝐷

𝐶𝐿

×

𝑐 𝜁𝜌

𝐶𝐿

×

[ln 𝑊0 − ln 𝑊1 ]

𝐶𝐷 𝜁𝜌

𝑅(𝑓𝑒𝑒𝑡) =

𝑐

𝐶𝐿

×

𝑊

[ln 𝑊0 ]

𝐶𝐷

1

𝜁𝜌

𝑅(𝑚𝑖𝑙𝑒𝑠) = 375 ×

𝐵𝑆𝐹𝐶

×

𝐶𝐿 𝐶𝐷

𝑊

×𝑙𝑛 𝑊0 1

𝐶𝐿

For best range, 𝐶 should be 𝐷

Breguet’s Formulas for Range and Endurance

maximum For the case of parabolic drag 𝐶𝐿

(𝐶 )

Range From eqn 2 𝑣𝑑𝑤 𝑑𝑠 = − 𝐶𝐵𝐻𝑃

𝐷

𝑣𝑑𝑤

𝑑𝑠 = − 𝑑𝑠 = − 𝑑𝑠 = −

𝐷𝑣 𝑐 𝜁𝜌

𝜁𝜌 𝑑𝑤 𝑐 𝐷 𝜁𝜌 𝐿

𝑙

×𝑤

×

𝑐 𝐷 𝜁𝜌 𝐶𝐿

𝑐 𝐶𝐷

1

=

2

𝑚𝑎𝑥

𝜋𝐴𝑒 𝜋𝐴𝑒

√𝐶

𝐷𝑜

𝑅𝑏𝑒𝑠𝑡 = 375×

At level flight, 𝑇𝐻𝑃𝐴𝑉 = 𝑇𝐻𝑃𝑟𝑒𝑞𝑑 𝐵𝐻𝑃. 𝜁𝜌 = 𝐷𝑉 𝐷𝑉 𝐵𝐻𝑃 = 𝜁𝜌 𝑑𝑠 = −

𝐶𝐿2

𝐶𝐷𝑜 = 𝐶𝐷𝑖 =

𝑑𝑤

𝑤 𝑑𝑤

×

𝑤

𝐷

Where: W0 = Initial gross weight W1 = Final aircraft weight (= W0-Wf) Wf = fuel weight used

𝐶

× (𝐶 𝐿 ) 𝐷

𝑊

𝑚𝑎𝑥

×𝑙𝑛 𝑊0 1

Endurance From eqn. ❶ 𝑑𝑤 𝑑𝑡 = − 𝐶𝐵𝐻𝑝

At level flight, 𝑇𝐻𝑃𝐴𝑣 = 𝑇𝐻𝑃𝑟𝑒𝑞𝑑 𝐷𝑣

𝐵𝐻𝑃 𝜁 𝑃 =

𝜁𝜌

Substitute 𝑑𝑤 𝑑𝑡 = − 𝐷𝑣 𝑐

𝑑𝑡 = − 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. 𝑓 𝑊1 𝑑𝑤 𝜁𝑝 𝐶 ∫𝑜 𝑑𝑠 = − 𝐶 (𝐶 𝐿 ) ∫𝑊2 𝑊

𝜁𝜌 𝐵𝑆𝐹𝐶

𝑑𝑡 = − 𝑑𝑡 = − 𝑑𝑡 = − 𝑑𝑡 = −

𝜁𝜌

𝜁𝜌 𝑐 𝜁𝜌 𝑐 𝜁𝜌 𝑐 𝜁𝜌 𝑐 𝜁𝜌 𝑐

3 𝐶𝐿2

×

𝑑𝑤 𝑣 𝐿

𝐿

×𝑤 𝑞

×𝐷×𝑣× 𝐶𝐿

×𝐶 × 𝐷

𝑑𝑤 𝑊 1

𝑤 2 1 √( )( )( ) 𝑠 𝜌 𝐶𝐿

𝐶

𝑠

𝜌

×

𝑑𝑤 𝑤

× 𝐶 𝐿 ×√(𝑤) ( 2) 𝐶𝐿 × ×

𝐷 3 𝐶𝐿 2

𝐶𝐷

𝜌𝑠

√2 ×

𝑑𝑤 𝑤

𝑑𝑤 3

𝑤2

𝜁𝜌, C, 𝐶 are all assumed to have 𝐷

constant

Average values throughout the flight,

Report No. 12 Range and Endurance

3

𝐸 ∫0 𝑑𝑡

E=

𝜁𝜌 𝑐

= − 3 𝐶𝐿2

𝜁𝜌 𝑐

𝐶𝐿2

1 𝑤1 −

𝜌𝑠

𝑤 3 −

𝐷

𝜁𝜌 𝑐

Given W= W0 = Wf= Wfused = Wl= W0 – Wfused s= CD=_____ + _____ CL2 (Clean Airplane) BHP= ̇ = 𝑊𝑓

0

𝐷

×𝐶 √2 × [

𝐸= 2

3

𝑊

𝜌𝑠

× 𝐶 √ 2 ∫𝑊 1 𝑊 −2 𝑑𝑤

3 𝐶𝐿2

1 3

] 𝑤0

𝜌𝑠

1

𝑤1

× 𝐶 √ 2 × [ 1] 𝐷

𝑤 2 𝑤0

3

E(seconds) =

𝜁𝜌 𝑐

𝐶𝐿2

× 𝐶 √2𝜌𝑠 ( 𝐷

1 √𝑤1

−

1 √𝑤0

)

3

𝐸(𝐻𝑜𝑢𝑟𝑠) = 778

𝜁𝜌 𝐵𝑆𝐹𝐶

×

𝐶𝐿2 𝐶𝐷

√𝜌𝑠 (

1

√𝑤1

−

1 √𝑤0

3

For the best endurance,

𝐶𝐿2

should

𝐶𝐷

be minimum in the case of parabolic drag polar equation, this yields: 𝐶𝐿2

1

𝐶𝐷𝑜 = 3 𝐶𝐷𝑖 =

3𝜋𝐴𝑒

𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷𝑜 𝐶𝐷 = 4𝐶𝐷𝑜 𝐸𝑏𝑒𝑠𝑡 (𝐻𝑜𝑢𝑟𝑠) = 778

𝜁𝜌 𝐵𝑆𝐹𝐶

3 𝐶𝐿2

( )

√𝜌𝑠 (

𝐶𝐷

√𝑤1

𝑚𝑎𝑥

Speed for best endurance 𝑤

2

−

1 √𝑤0

)

Solution 𝑊̇

𝑓 𝐵𝑆𝐹𝐶 = 𝐵𝐻𝑃 𝐵𝑆𝐹𝐶𝐶𝐿 = 75 ℎ 𝐵𝑆𝐹𝐶

Best Range, Rbest : 𝜁𝜌 𝐶 𝑅𝑏𝑒𝑠𝑡 (𝑚𝑖𝑙𝑒𝑠) = 375 𝐵𝑆𝐹𝐶 (𝐶 𝐿 ) 𝐶𝑟

𝐷

𝑤

𝑚𝑎𝑥

Where: 𝐶𝐿 = √𝜋𝐴𝑒𝐶𝐷𝑜 𝐶𝐷 = 2𝐶𝐷𝑜

𝑙𝑛 𝑤0

𝑓

1

𝑉𝐸𝑏𝑒𝑠𝑡 = √( 𝑠 ) (𝜌) (𝐶 ) 𝐿

Where: 𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷𝑜

1

)

Calculate: (a) Rbest and corresponding flight speed (b) Ebest and corresponding flight speed

Flight Speed for Rbest , VRBest : 𝑤

2

1

𝑉𝑅𝐵𝑒𝑠𝑡 = √( 𝑠 ) (𝜌) (𝐶 ) 𝐿

Where: 𝐶𝐿 = √𝜋𝐴𝑒𝐶𝐷𝑜 For Best Endurance, Ebest,: 3

𝐸𝑏𝑒𝑠𝑡 (𝐻𝑜𝑢𝑟𝑠) = 778

𝜁𝜌 𝐵𝑆𝐹𝐶

Where: 𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷𝑜 𝐶𝐷 = 4𝐶𝐷𝑜

𝐶𝐿2

( )

√𝜌𝑠 (

𝐶𝐷

1

√𝑤1

𝑚𝑎𝑥

−

1 √𝑤0

)

Flight Speed for Best Endurance, VEBest : 𝑤

2

1

𝑉𝐸𝐵𝑒𝑠𝑡 = √( 𝑠 ) (𝜌) (𝐶 ) 𝐿

Where: 𝐶𝐿 = √3𝜋𝐴𝑒𝐶𝐷𝑜

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 ζ𝜌= 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 BSFC = 0.45 lb CD = 0.02 + 0.05 CL2 τρ = 0.87

1

CL = √(0.05) (0.02) CL = 0.632 CD = 2 CD = 2(0.02) = 0.04 30000

s = 300 ft2

𝑊𝐼 =

h = 28 000 ft

𝑊𝐼 = 26 317.92 lb

1500 0.45 0.04 [( )( )( )] 𝑒 375 0.87 0.632

Wsupply = 3000 lb R = 1500 miles

Amount of fuel consumed Wf used = Wc -W𝐼

Required:

= 30000 – 26317.92

Total amount of fuel consumed, Wf used(total), and corresponding flying time, Etotal

Wf used = 3682.08 lb

Initial Flight Speed Solution:

𝑊

At h = 28000 ft

a) To destination τρ

𝐶

𝐿

ln

Wc Wf

𝑊𝑜

=𝑒 𝑊

𝑅

𝐵𝑆𝐹𝐶

= (375) ( [(

τρ

𝜌 = 𝜌𝑜 (1 +

Wc W𝐼 𝐶

) ( 𝐶𝐷 ) 𝐿

𝑅 𝐵𝑆𝐹𝐶 𝐶𝐷 )( )( )] 375 τρ 𝐶𝐿

𝐼

𝑊𝐼 =

Wo

1

𝐿

The computation will be done in two parts:

R = 375 𝐵𝑆𝐹𝐶 ( 𝐶𝐷) ln

2

V= √( 𝑠 ) (𝜌) (𝐶 )

𝜌 = 0.002377 (1 +

𝑎ℎ 𝑇𝑜

)

(−0.003566)(28000) 4.26 ) 519

𝜌 = 0.000957 slug/ft3 V= √(

30000 300

2

1

) (0.000957) (0.632)

V= 575.04 fps

𝑅 𝐵𝑆𝐹𝐶 𝐶𝐷 [( )( )( )] τρ 𝐶𝐿 𝑒 375

Flying time For optimum range, a/ CD should be kept at maximum, CL = √𝜋𝐴𝑒𝐶𝐷

3

τρ 𝐶𝐿 2

E = 778 𝐵𝑆𝐹𝐶

𝐶𝐷

1

1

√𝜌𝑠 (√𝑊 − √𝑊 ) 𝐼

𝑜

3

(0.87) (0.632)2

E = 778 (

0.45

1 √26317.92

−

0.04 1

√(0.000957)(300)

)

√30000

Total Flying Time

b) Return Trip

Etotal = 3.95 + 4.49

Wc = 26317.92 -3000

Etotal = 8.44 hours

Wc = 23317.92 lb

𝑊𝐼 =

Wo 𝑅 𝐵𝑆𝐹𝐶 𝐶𝐷 [( )( )( )] τρ 𝐶𝐿 𝑒 375

Problem 11.1 Determine the maximum range, maximum endurance (and speeds for best range and endurance at 10000 ft) of the following airplane:

23317.92 1500 0.45 0.04 [( )( )( )] 𝑒 375 0.87 0.632

𝑊𝐼 = 20445.97 lb

S = 200 ft2

Amount of Fuel Consumed

W = 10000 lb

Wf used = Wc -W𝐼 = 23317.92 – 20455.97 Wf used = 2861.95 lb

τρ

E = 778 𝐵𝑆𝐹𝐶 E = 778 1 √20455.97

3 𝐶𝐿 2

𝐶𝐷

(0.87)

𝐼

0.04 1

)

√23317.92

E = 4.49 hours

1

√𝜌𝑠 (√𝑊 − √𝑊 )

3 (0.632)2

0.45

−

1

Wf = 4000 lb BSFC = 0.52 lb/BHP-hr Τρ = 0.90

Flying time

(

Wf used(total) = 3682.08 + 2861.95 Wf used(total) = 6544.03 lb

E = 3.95 hours

𝑊𝐼 =

Total Amount of Fuel Consumed

𝑜

√(0.000957)(300)

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 𝐵𝑆𝐹𝐶 ( 𝐶𝐷 ) ln

130

220

125

240

τρ

𝐶

𝐿

Wc W𝐼

3

τρ 𝐶𝐿 2

E(hours) = 778 𝐵𝑆𝐹𝐶 1 √𝑊𝑜

𝐶𝐷

1

√𝜌𝑠 (√𝑊 − 𝐼

)

Req’d: Vbest and VRbest Ebest and VEbest

Solution: 𝑎ℎ

𝜌 = 𝜌𝑜 (1 +

𝑇𝑜

)

𝜌 = 0.002377 (1 +

(−0.003566)(10000) 4.26 ) 519

𝜌 = 0.001755 slug/ft3

CL = 1 2

𝑊

Centripetal Force 1. A plane of 3800 lb gross weight is turning at 175 mph with an angle of bank of 50˚. (a)What is the centrifugal force? (b)What is the lift? (c)What would be the radius of turn? Given:

22

V = 175 mph X 15

𝜌𝑣 2 𝑠 10000

=1

= 256.67 ft/sec

22 2 (0.001755)(9𝑋 ) (200) 2 15

CL =

26486.66 𝑣2

, where v is in graph

𝐷𝑉

THPreq’d = 375 D=

375THPreq’d 𝑉

D = lbs V = mph 𝐶

D = ( 𝐶𝐷 ) 𝑊 𝐿

𝐶𝐷 =

𝐶𝐿 𝐷 𝑊

W = 3800 lb

Β = 50˚ Req’d: a) CF b) L c) R Sol’n: a) tan β =

𝐶𝐹 𝑊

CF = W tan β CF = (3800)( tan 50˚) CF = 4528.66 lb

Minimum Speed in Turns

𝑊

b) L = cos 𝛽 3800

= cos 50˚ L= 5911.75 lb

c) tan β = R= R=

𝑣2 𝑔𝑅

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?

𝑣2

Given: Vs = 39.3 mph

tan βg

Req’d: a) Vs @ β = 30˚

256.612

b) Vs @ β = 50˚

tan 50˚(32.174)

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˚ R =565 ft Req’d: v

Vs =

a) For Vs @ 30˚ Vs =

tan β =

𝑣2 𝑔𝑅

v2 = gRtan𝛽 v = √gRtan𝛽 v = √(32.174)(565)(tan 40˚) v = 123.50 ft/sec

39.3 √cos 30˚

= 42.23 mph

b) For Vs @ 50˚ Vs =

Sol’n :

𝑉𝑠 √cos 𝛽

39.3 √cos 50˚

= 49.03 mph

c) For Vs @ 70˚ Vs =

39.3 √cos 70˚

= 67.20 mph

2. An airplane with a loading of 18.2 lb/ft2 uses a wing section whose CLmax is 1.5. What is the stalling speed in a 40˚ banked turn at sea level conditions? Given:

𝑤 𝑠

= 18.2 lb/ft2

CLmax = 1.5 Β = 40˚ Req’d: Vs Sol’n: 𝑤

2

Vs = ( ) ( ) ( 𝑠 ρ 𝐶

1

𝐿𝑚𝑎𝑥

2

1

) (cos 𝛽) 1

1

= (18.2) (0.002377) (1.5) (cos 40) Vs = 13326.83