Appendix 10.2

AN EXAMPLE OF AIRPLANE

PRELIMINARY DESIGN

PROCEDURE - JET TRANSPORT

E.G.Tulapurkara

A.Venkattraman

V.Ganesh

REPORT NO: AE TR 2007-4

APRIL 2007

An Example of Airplane Preliminary DesignProcedure - Jet Transport

E.G.Tulapurkara∗

A.Venkattraman†

V.Ganesh‡

Abstract

In this report, we present an application of the preliminary designprocedure followed in aircraft design course. A 150 seater jet airplanecruising at M = 0.8, at 11 km altitude and having a gross still airrange(GSAR) of 4000 km is considered. The presentation is dividedinto eight sections

• Data collection

• Preliminary Weight estimation

• Optimization of wing loading and thrust loading

• Wing design

• Fuselage design, preliminary design of tail surface and prelimi-nary layout

• c.g. calculation

• Control surface design

• Features of designed airplane

• Details of performance estimation

∗AICTE Emeritus Fellow, Department of Aerospace Engineering, IIT Madras†B.Tech Student, Department of Aerospace Engineering, IIT Madras‡Dual Degree Student, Department of Aerospace Engineering, IIT Madras

1

Contents

1 Data Collection 61.1 The Design Philosophy . . . . . . . . . . . . . . . . . . . . . . 6

1.1.1 Type of Aircraft and Market . . . . . . . . . . . . . . . 61.1.2 Budget and Time Constraints . . . . . . . . . . . . . . 71.1.3 Other Constraints and Standards . . . . . . . . . . . . 7

1.2 Preliminary Design . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 Preliminary Weight Estimate . . . . . . . . . . . . . . 91.2.2 Wing parameters . . . . . . . . . . . . . . . . . . . . . 91.2.3 Empennage . . . . . . . . . . . . . . . . . . . . . . . . 101.2.4 Control Surfaces . . . . . . . . . . . . . . . . . . . . . 111.2.5 Fuselage . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.6 Engines . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.7 Landing Gear . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Overall height . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Revised Weight Estimation 212.1 Fuel fraction estimation . . . . . . . . . . . . . . . . . . . . . 21

2.1.1 Warm up and Take off . . . . . . . . . . . . . . . . . . 212.1.2 Climb . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1.3 Cruise . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.4 Loiter . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.1.5 Landing . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2 Empty Weight Fraction . . . . . . . . . . . . . . . . . . . . . . 23

3 Wing Loading and Thrust Loading 253.1 Landing Distance Consideration . . . . . . . . . . . . . . . . . 253.2 Maximum Speed(Vmax) Consideration . . . . . . . . . . . . . . 27

3.2.1 Estimation of K . . . . . . . . . . . . . . . . . . . . . . 273.3 (R/C)max consideration . . . . . . . . . . . . . . . . . . . . . 323.4 Based on Minimum Fuel for Range (Wfmin

) . . . . . . . . . . . 333.5 Based on Absolute Ceiling . . . . . . . . . . . . . . . . . . . . 343.6 Summary of Constraints . . . . . . . . . . . . . . . . . . . . . 363.7 Consideration of Wing Weight (Ww) . . . . . . . . . . . . . . 363.8 Choosing a W/S . . . . . . . . . . . . . . . . . . . . . . . . . 373.9 Thrust Requirements . . . . . . . . . . . . . . . . . . . . . . . 37

3.9.1 Requirement for Vmax . . . . . . . . . . . . . . . . . . . 383.10 Requirements for (R/C)max . . . . . . . . . . . . . . . . . . . 383.11 Take-Off Thrust Requirements . . . . . . . . . . . . . . . . . . 383.12 Engine Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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3.13 Engine Characteristics . . . . . . . . . . . . . . . . . . . . . . 39

4 Wing Design 424.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.2 Airfoil Selection . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2.1 Design Lift Coefficient . . . . . . . . . . . . . . . . . . 434.2.2 Airfoil Thickness Ratio and Wing Sweep . . . . . . . . 43

4.3 Other Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 444.3.1 Aspect Ratio . . . . . . . . . . . . . . . . . . . . . . . 444.3.2 Taper Ratio . . . . . . . . . . . . . . . . . . . . . . . . 454.3.3 Root and Tip Chords . . . . . . . . . . . . . . . . . . . 454.3.4 Dihedral . . . . . . . . . . . . . . . . . . . . . . . . . . 454.3.5 Wing Twist . . . . . . . . . . . . . . . . . . . . . . . . 45

4.4 Cranked Wing Design . . . . . . . . . . . . . . . . . . . . . . 464.5 Wing Incidence(iw) . . . . . . . . . . . . . . . . . . . . . . . . 474.6 Vertical Location of Wing . . . . . . . . . . . . . . . . . . . . 474.7 Areas of Flaps and Ailerons . . . . . . . . . . . . . . . . . . . 48

5 Fuselage and Tail Layout 485.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485.2 Initial Estimate of Fuselage Length . . . . . . . . . . . . . . . 485.3 Nose and Cockpit - Front Fuselage . . . . . . . . . . . . . . . 495.4 Passenger Cabin Layout . . . . . . . . . . . . . . . . . . . . . 49

5.4.1 Cabin Cross Section . . . . . . . . . . . . . . . . . . . 505.4.2 Cabin length . . . . . . . . . . . . . . . . . . . . . . . 505.4.3 Cabin Diameter . . . . . . . . . . . . . . . . . . . . . . 51

5.5 Rear Fuselage . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.6 Total Fuselage Length . . . . . . . . . . . . . . . . . . . . . . 515.7 Tail surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.8 Engine Location . . . . . . . . . . . . . . . . . . . . . . . . . . 545.9 Landing Gear Arrangement . . . . . . . . . . . . . . . . . . . 54

6 Estimation of Component Weights and C.G Location 556.1 Aircraft mass statement . . . . . . . . . . . . . . . . . . . . . 55

6.1.1 Structures Group . . . . . . . . . . . . . . . . . . . . . 556.1.2 Propulsion Group . . . . . . . . . . . . . . . . . . . . . 566.1.3 Fixed equipment group . . . . . . . . . . . . . . . . . . 56

6.2 Weights of Various Components . . . . . . . . . . . . . . . . . 576.3 C.G Location and C.G Travel . . . . . . . . . . . . . . . . . . 57

6.3.1 Wing Location on Fuselage . . . . . . . . . . . . . . . . 576.4 C.G Travel for Critical Cases . . . . . . . . . . . . . . . . . . 58

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6.4.1 Full Payload and No Fuel . . . . . . . . . . . . . . . . 586.4.2 No Payload and No Fuel . . . . . . . . . . . . . . . . . 586.4.3 No Payload and Full fuel . . . . . . . . . . . . . . . . . 596.4.4 Payload distribution for 15% c.g travel . . . . . . . . . 59

6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7 Control Surfaces 607.1 Stability and Controllability . . . . . . . . . . . . . . . . . . . 607.2 Static Longitudinal Stability and Control . . . . . . . . . . . . 60

7.2.1 Specifications . . . . . . . . . . . . . . . . . . . . . . . 607.2.2 Aft Center of gravity limit . . . . . . . . . . . . . . . . 607.2.3 Forward center of Gravity Limit . . . . . . . . . . . . . 617.2.4 Determination of initial parameters . . . . . . . . . . . 61

7.3 Lateral Stability and Control . . . . . . . . . . . . . . . . . . 657.3.1 Specifications . . . . . . . . . . . . . . . . . . . . . . . 657.3.2 Equations for directional stability . . . . . . . . . . . . 657.3.3 Determination of initial parameters . . . . . . . . . . . 65

8 Features of the Designed Airplane 678.1 Three View Drawing . . . . . . . . . . . . . . . . . . . . . . . 678.2 Overall Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 678.3 Engine details . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.4 Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678.5 Wing Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 698.6 Fuselage Geometry . . . . . . . . . . . . . . . . . . . . . . . . 698.7 Nacelle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 698.8 Horizontal Tail Geometry . . . . . . . . . . . . . . . . . . . . 698.9 Vertical Tail Geometry . . . . . . . . . . . . . . . . . . . . . . 708.10 Other details . . . . . . . . . . . . . . . . . . . . . . . . . . . 708.11 Crew and Payload . . . . . . . . . . . . . . . . . . . . . . . . 708.12 Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

9 Performance Estimation 729.1 Estimation of Drag Polar . . . . . . . . . . . . . . . . . . . . . 72

9.1.1 Estimation of (CDo)WB . . . . . . . . . . . . . . . . . . 729.1.2 Estimation of (CDo)V and (CDo)H . . . . . . . . . . . 749.1.3 Estimation of Misc Drag - Nacelle . . . . . . . . . . . . 759.1.4 CDo of the airplane . . . . . . . . . . . . . . . . . . . . 759.1.5 Induced Drag . . . . . . . . . . . . . . . . . . . . . . . 759.1.6 Final Drag Polar . . . . . . . . . . . . . . . . . . . . . 76

9.2 Engine Characteristics . . . . . . . . . . . . . . . . . . . . . . 77

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9.3 Level Flight Performance . . . . . . . . . . . . . . . . . . . . . 809.3.1 Stalling speed . . . . . . . . . . . . . . . . . . . . . . . 809.3.2 Variation of Vmin and Vmax with Altitude . . . . . . . . 82

9.4 Steady Climb . . . . . . . . . . . . . . . . . . . . . . . . . . . 909.5 Range and Endurance . . . . . . . . . . . . . . . . . . . . . . 969.6 Turning Performance . . . . . . . . . . . . . . . . . . . . . . . 999.7 Take-off distance . . . . . . . . . . . . . . . . . . . . . . . . . 1039.8 Landing distance . . . . . . . . . . . . . . . . . . . . . . . . . 1049.9 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . 105

10 Acknowledgements 107

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1 Data Collection

1.1 The Design Philosophy

The conceptual design forms the initial stage of the design process. In spiteof the fact that there are numerous aircrafts, each having its own special fea-tures, one can find common features underlying most of them. For example,the following aspects would dominate the conceptual design of a commercialtransport jet.

1.1.1 Type of Aircraft and Market

The Civil Transport Jets could be classified in the following way :

Class No.of Seats Typical GSAR(km) PropulsionB-747 >400 >13000 High Bypasstype Turbofan

B-757 200-400 10000 High Bypasstype Turbofan

B-737 100-200 5000 Medium Bypasstype Turbofan

Regionals 30-100 2000 Turboprop

Table 1: Classification of Civil Jet Airplane

From the values of gross still air range in table, it is clear that inter-continental flights would be restricted to the first two classes while the lasttwo would handle bulk of the traffic in regional routes. The different classescater to different sections of the market. One decides the range and pay-load(ie passengers) after identifying the target market. In this example, weplan to cater to the traffic in regional routes. We will design a TransportJet with a Gross Still Air Range(GSAR) of 4000km (=Rg) and a single-classseating capacity of 150. We could roughly classify our aircraft as belongingthe B-737 class. We collect data for similar aircrafts and use this data set asthe basis for making initial estimates.

Our aim is to design an aircraft that satisfies the following requirements.

• Gross Still Air Range = 4,000 km

• No. of passengers = 150

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• Flight Cruise Mach No. = 0.80

• Altitude =11,000 m

1.1.2 Budget and Time Constraints

Any design team would be required to work with a limited amount of fundsand time. These could dictate various aspects of the design process.For exam-ple, innovations which could end up in a spiralling budget may be shelved.Also, in case of highly competitive markets, the ability to get the aircraftready in the prescribed time frame is very crucial. The design team mustensure that cost and time over-runs are minimized to the extent possible.

1.1.3 Other Constraints and Standards

Some of the major demands on the design arise from the various mandatoryand operational regulations. All commercial aircrafts must satisfy the Air-worthiness requirements of various countries. Typically, each country has itsown Aviation Authority (e.g, DGCA in India, CAA in UK, FAA in USA).Airworthiness requirements would cover the following aspects of the aircraft

1. FlightThis includes performance items like stall, take-off, climb, cruise, de-scent, landing, response to rough air etc. Also included are require-ments of stability,controllability and manoeuvrability.

2. StructuralFlight loads, ground loads, emergency landing conditions, fatigue eval-uation etc.

3. PowerplantFire protection, auxillary power unit,air intake/exhaust,fuel systems,cooling.

4. OtherMaterials quality regulations, bird strike.

Passenger Safety is the primary motive behind these specifications. Ad-ditional route-specific constraints may have to be taken into account on acase-by-case basis. e.g, cruise altitude for aircrafts flying over the Himalayasmust be well over 8 km.

In addition to safety and operational requirements, the design must satisfythe environmental constraints. Two major environmental concerns are noiseand emissions :

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• The Engines are the primary source of noise in an aircraft. The airframecould also add to this.Maximum noise is produced during take-off andlanding. This can reduced by opting for a shallower approach, as thisreduces the flight time spent near the airport. However the reduction innoise may not be significant. The development of high-bypass turbofanengines has significantly reduced noise production.

• The predominant source of emissions is the engine. The exhaust con-tains particles, various gases including carbon dioxide(CO2) , water va-por (H2O) , various oxides of nitrates, carbon monoxide(CO),unburnthydrocarbons and sulphur dioxide(SO2). All components except CO2

and H2O are considered as pollutants Again,as was the case with noise,emissions during landing and take-off are of particular concern due tothe communities near airports. Various aviation authorities have setlimits on these emissions. The design team must adhere to such con-straints.

1.2 Preliminary Design

If we look at the commercial transport jets in use, one can find many commonfeatures amongst them. Some of these are :

• Medium bypass turbofansThis choice regarding the type of engine is due to the following reasons.In the flight regime of Mach number between 0.6 to 0.85, turbofans givethe best efficiency and moreover reduction in thrust output with speedis not so rapid. Also, the noise generated by a medium-by pass turbofan engine is considerably less. We follow this trend and choose amedium-by pass turbo fan as our powerplant.

• Wing mounted engines Though not a rule, wing mounted enginesdominate the designs of top aircraft companies like Boeing and Airbus.Alternative designs could be adopted. But,given the experience gainedwith the wing mounted engines and the large data available for suchconfigurations, we adopt two wing mounted engines.

• Swept back wings and a conventional rear-tail configuration is cho-sen. Again, this choice is dictated by the fact that we have a largeamount of data(to compare with) for such configurations.

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1.2.1 Preliminary Weight Estimate

Given the number of passengers, we can estimate the payload in the followingway:

1. Include one cabin crew member for 30 passengers. In our case, thisgives 5 crew members

2. Include flight crew of pilot and co-Pilot.Thus the total of passenger + crew is 150+5+2 = 157.

3. Allow 110 kg for each passenger (82 kg weight per passenger with carryon baggage + 28 kg of checkin baggage)(Reference 1.11, page 214)

We thus obtain a payload Wpay of 157× 110 = 17270 kgf . We now esti-mate the gross weight of the aircraft (Wg).

From data collection, we observe the following.

Aircraft No.of passengers Still air range (km) WTO(kgf)737-300B 149 4185 60636737-400B 168 3852 64671737-700A 149 2935 60330

Table 2: Take off weight

Based on the data collected, we choose an initial weight of 60,000 kgf .

1.2.2 Wing parameters

To estimate the wing parameters, we need to choose a value for wing loading(W/S).This is one of the most important parameters that not only decides the wingparameters but also plays an important role in the performance of the air-plane.We observe similar airplanes and choose an initial estimate for (W/S)to be 5500 N/m2.Once the (W/S) has been decided, the other parametersof the wing are chosen based on similar aircraft.

Aerodynamically, it is desirable to have a large aspect ratio(A). How-ever, structural considerations force us to settle for an optimal value. As thestructural design improves, the value of A also keeps increasing. We choosea value of 9.3. Most modern aircrafts(see data base in Table A) have valuesclose to 9.The taper ratio(λ) is a geometric parameter that is roughly the

9

same for all the aircrafts in the data set. We choose an average value of 0.24for λ.The wing quarter chord sweep(Λc/4) is chosen as 25◦.Consequently

S = Wg

(S

W

)= 107.02m2 (1)

The wing span(b) can be calculated from A and S

b =√SA = 31.55 m (2)

The root chord(cr) and tip chord(ct) can now be found using the followingequations :

cr =2S

b(1 + λ)= 5.47 m (3)

ct = λcr = 1.31 m (4)

1.2.3 Empennage

As explained earlier,we have chosen the conventional rear-tail configuration.The geometric parameters of the horizontal and vertical tails are obtainedhere.

The values of Sh/S and Sv/S are obtained from the data set of similarairplanes.

We have chosen

ShS

= 0.31

SvS

= 0.21

Hence,

Sh = 33.18 m2

Sv = 22.47 m2

We choose suitable aspect ratios(Ah, Av) from the data set. Our choicesare Ah = 5 and Av = 1.7. Using eq.(2), we get the spans(bh, bv) as

bh =√AhSh = 12.88 m (5)

10

bv =√AvSv = 6.18 m (6)

The chosen values for the taper ratios(λh, λv) from the data set are λh =0.26 λv = 0.3. We can now compute the root chord (crh, crv) and tip chord(cth, ctv) of tails as

crh =2Sh

bh(1 + λh)= 4.09 m (7)

cth = λhcrh = 1.06 m (8)

crv =2Sv

bv(1 + λv)= 5.59 m (9)

ctv = λvcrv = 1.68 m (10)

From the data set, we choose quarter chord sweep back angles of Λh = 30◦

and Λv = 35◦. This completes the broad geometric design of the empennage.

1.2.4 Control Surfaces

A number of aircraft and their 3-view drawings as well as design data havebeen studied and the following parameter values are chosen.

• Sflap/S = 0.17

• Sslat/S = 0.10

• bflap/b =0.74

• Sele/Sht = 0.22

• Srud/Svt = 0.25

• Trailing edge flaps type : Fowler flaps

• Leading edge high lift devices : slats

Hence,

• Sele = 7.53 m2

• Srud = 5.8 m2

• Area of T.E flaps = 18.98 m2

• Area of L.E slats = 11.60 m2

• bflap = 23.7 m

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1.2.5 Fuselage

Aerodynamic considerations would demand a slender fuselage. But, pas-senger comfort and structural constraints would limit the slenderness. Weobtain the length lf and diameter df by choosing lf/b = 1.05 and lf/df =8.86 from data collection.

Hence,

lf = 33.6 m (11)

df = 3.79 m (12)

1.2.6 Engines

Observing the thrust-to-weight ratio (T/W) of similar airplanes, we arriveat a T/W of 0.3.This implies a thrust requirement of

T = 0.3×Wg = 180 kN or 90 kN per engine

The CFMI FM56-3-B1 model of Turbofan comes closest to this re-quirement.

1.2.7 Landing Gear

We choose a retractable tricycle type landing gear. It is the most commonlyfound type of landing gear. It is favored for two reasons:

1. During take-off and landing the weight of the plane is taken entirely bythe rear wheels.

2. It has better lateral stability on ground than bicycle type landing gear.

We choose to have a total of 10 wheels - 2 below the nose and two pairseach on the sides(near the wing fuselage junction). The location of the wheelswas chosen from three-view drawings of similar aircraft.

1.3 Overall height

Based on dimensions of Boeing 737 - 300, 400 and 500, the overall height istaken as 11.13 m.

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TABLE A - Data on Existing Airplanes(150 seater category)(Source : http://www.bh.com/companions/034074152X/)

13

14

15

16

17

Figure 1: Three view drawing of Boeing 737-300

Source : http://www.virtualswa.com/Boeing737-300/3view.gif

Figure 2: Three view drawing of Boeing 737-500

Source : http://www.virtualswa.com/Boeing737-500/3view.gif

18

Figure 3: Three view drawing of Boeing 737-700

Source : http://www.virtualswa.com/Boeing737-700/3view.gif

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Fig

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ree

vie

wof

the

airp

lane

under

des

ign

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2 Revised Weight Estimation

In the previous section, an initial estimate for the aircraft parameters hasbeen done. The weight estimate is being revised using refined estimatesof fuel weight and empty weight. The fuel fractions for various phases areworked out in the following steps. The fuel fractions for warm-up, take-off,climb and landing are taken from Raymer[4], chapter 3.

2.1 Fuel fraction estimation

The fuel weight depends on the mission profile and the fuel required as re-serve. The mission profile for a civil jet transport aircraft involves

• Take off

• Climb

• Cruise

• Loiter before landing

• Descent

• Landing

2.1.1 Warm up and Take off

The value for this stage is taken by following the standards given in Raymer[4],chapter 3

W1

W0

= 0.97

W0 is the weight at take-off and W1 is the weight at the end of the take-offphase.

2.1.2 Climb

The weight-ratio for this stage is chosen by following the standards given inRaymer[4], chapter 3.

W2

W1

= 0.985

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2.1.3 Cruise

The weight ratio for the cruise phase of flight is calculated using the followingexpression from Raymer[4], chapter 3.

W3

W2

= exp

[−RCV (L/D)

](13)

Gross still air range is 4000 km.Hence

Cruise Safe Range =GSAR

1.5=

4000

1.5= 2667 km

(L/D)max is taken as 18 from figure 3.6 of Raymer[4]. This correspondsto the average value for civil jets.

As prescribed by Raymer[4], chapter 3

(L/D)cruise = 0.866(L/D)max (14)

(L/D)cruise = 0.866× 18 = 15.54

To account for allowances due to head wind during cruise and provisionfor diversion to another airport we proceed as follows.

Head wind is taken as 15 m/s. The time to cover the cruise safe range of2667 km at Vcr of 849.6 km/hr is

Time =2667

849.6= 3.13 hours

Therefore, with a head wind of 15 m/s or 54 km/hr the additional dis-tance that has to be accounted for is

Additional distance = 54× 3.13 = 169 km

The allowance for diversion to another airport is taken as 400 km.The total extra distance that has to be accounted for in the calculations

is 169 + 400 = 569 km.

The total distance during cruise = 2667 + 569 = 3236 km.Substituting the appropriate values in eq.(13) we get,

W3

W2

= exp

[−3236× 0.6

849.6× 15.59

]= 0.863

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2.1.4 Loiter

The weight ratio for Loiter phase of flight is calculated using the followingexpression from Raymer[4], chapter 3

W4

W3

= exp

[−E × TSFC

(L/D)

](15)

During Loiter, the airplane usually operates at (L/D)max and hence theappropriate value should be used in eq.(15). Also, we design for a loiter timeof 30 minutes.

Therefore we get,

W4

W3

= exp

[−0.5× 0.6

18

]= 0.983

2.1.5 Landing

Following the standards specified by Raymer[4], chapter 3, we take this ratioas

W5

W4

= 0.995

Therefore,

W5

Wg

=W5

W0

= 0.97× 0.985× 0.863× 0.983× 0.995 = 0.806

Allowing for a reserve fuel of 6% we obtain the fuel fraction(ζ) as

Wf

Wg

= ζ = 1.06

(1− W5

W0

)= 0.205

2.2 Empty Weight Fraction

To determine the empty weight ratio, we follow the method in Raymer[4],chapter 3 which gives a relation between We/Wg and Wg as follows.

We

Wg

= 1.02(2.202Wg)−0.06 (16)

where Wg is in kgf .

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Hence,

Wg =Wpay

1−Wf/Wg −We/Wg

=17270

1− 0.205− 1.202(2.202Wg)−0.06(16A)

We solve this equation by iteration

Wg(guess) We/Wg(from eq.(16)) Wg(from eq.(16A))60000 0.50274 5909059090 0.50320 5918459184 0.50315 5917459174 0.50316 5917559175 0.50316 59175

Table 3: Iterative procedure for Wg

Hence, the gross weight Wg is obtained as

Wg = 59, 175 kgf

The critical weight ratios are

We

Wg

= 0.503

Wf

Wg

= 0.205

Wpay

Wg

= 0.292

24

3 Wing Loading and Thrust Loading

The thrust-to-weight ratio (T/W ) and the wing loading(W/S) are the twomost important parameters affecting aircraft performance. Optimization ofthese parameters forms a major part of the design activities conducted afterinitial weight estimation. For example, if the wing loading used for the initiallayout is low, then the area would be large and there would be enough spacefor the landing gear and fuel tanks. However it results in a heavier wing.

Wing loading and thrust-to-weight ratio are interconnected for a numberof critical performance items, such as take-off distance, maximum speed etc.These are often the design drivers. A requirement for short takeoff can bemet by using a large wing (low W/S) with a relatively low T/W . On theother hand, the same takeoff distance could be met with a high W/S alongwith a higher T/W .

In this section, we use different criteria and optimize the wing loadingand thrust loading.

Wing loading affects stalling speed, climb rate,takeoff and landing dis-tances, minimum fuel required and turn performance.

Similarly, a higher thrust loading would result in more cost which is un-desirable. However it would also lead to enhanced climb performance.

Hence a trade-off is needed while choosing W/S and T/W . Optimizationof W/S and T/W based on various considerations is carried out in the fol-lowing subsections.

3.1 Landing Distance Consideration

To decide the wing loading from landing distance consideration we needto choose the landing field length. Based on data collection of similar air-craft(Table A) the landing field length is chosen to be 1425 m.

sLand = 1425 m

Next,we choose the CLmax of the airplane. The Maximum lift coefficientdepends upon the wing geometry,airfoil shape,flap geometry and span,leadingedge slot or slat geometry,Reynolds number,surface texture and interference

25

from other parts of the aircraft such as the fuselage,nacelles or pylons.

Raymer[4], chapter 5 provides a chart for CLmax as a function of Λc/4 fordifferent types of high lift devices(figure 5.3 of Raymer[4]). For our airplanewe decided to use Fowler flap and slat as the high lift devices. This gives usa CLmax of 2.5 for a Λc/4 = 25o.

CLmax = 2.5

To calculate W/S based on landing considerations,we use the formula

W

S=

1

2ρV 2

s CLmax (17)

The stalling speed Vs is estimated in the following way,

sLand = 1425 m

The approach speed (Va) in knots is related to the landing distance(sLand)in feet as,

Va(in knots) =

√sLand0.3

= 128.34 knots = 64.17 ms−1

From the approach speed, the stalling speed can be calculated,

Vs =Va0.3

= 49 ms−1 (18)

Now, using this value for Vs in eq.(17),(W

S

)Land

= 3743 Nm−2

Since WLand = 0.85Wt.o the W/S at take-off is,(W

S

)t.o

=1

0.85

(W

S

)Land

= 4403 Nm−2

Allowing a 10 % variation in Vs we get a range of wing loading as

3639 < p < 5328 N/m2

26

3.2 Maximum Speed(Vmax) Consideration

Generally the Mmax is determined as follows

Mmax = Mcr + 0.04

Hence,for our airplane,

Mmax = 0.80 + 0.04 = 0.84

The drag polar is generally expressed as

CD = CD0 +KC2L (19)

where,

K =1

πAe(20)

CD0 for the airplane is given as

CD0 = Cfe ×SwetS

(21)

Swet/S = 6.33 from Fig 2.5 of Raymer[4].

3.2.1 Estimation of K

We estimate ‘e’ from Roskam[6], chapter 2

1

e=

1

ewing+

1

efuse+ 0.05 (22)

ewing = 0.84 for unswept wing of A = 9.3 and λ = 0.25.Hence,ewing for the swept wing is

ewing = 0.84 cos(Λ− 5) = 0.84 cos(25− 5) = 0.7893 (23)

1

efuse= 0.1

Hence,

1

e=

1

0.7893+ 0.1 + 0.05 = 1.417

e = 0.707

27

K =1

π × 9.3× 0.707= 0.0482

To get CD0 we note from figure 3.6 of Raymer[4] that (L/D)max=18.Thishas already been used in section 2.

(L/D)max =1

2√CD0K

(24)

Hence,

CD0 =1

4K(L/D)2max

=1

4× 0.0482× 182= 0.0161

Further,

CD0 = CfeSwetS

(25)

gives,

Cfe =0.0161

6.33= 0.00254

Hence, the drag polar is

CD0 = 0.0161 + 0.0482C2L

To obtain the optimum W/S based on maximum speed,we the followmethod given in Lebedinski[7], chapter IV of writing the drag polar as afunction of p (=W/S)

CD = F1 + F2p+ F3p2 (26)

where,

F1 = Cfe

(1 +

ShtS

+SvtS

) (SwetS

)w

= CfeKt (27)

F2 =(CDo − F1)

W/S(28)

F3 =K

q2(29)

To calculate F1, F2, F3 values for our airplane we proceed as follows.

28

From our preliminary estimations ,

ShtS

= 0.31

SvtS

= 0.21

Hence,

Kt = 1 +ShtS

+SvtS

= 1.52

(CDo

)W

= Cfe

(Swet(exposed)

S

)W

(30)

To calculate (Swet(exposed)/S)W we need to obtain dimensions of the ex-posed wing.We proceed as follows. From preliminary estimate in section 1

• S = 107.02 m2

• λ = 0.24

• A = 9.3

• cr = 5.47 m

• ct = 1.31 m

• Λc/4 = 25◦

Hence, for the equivalent trapezoidal wing, the chord distribution is givenby

c(y) = cr −cr − ctb/2

y

= 5.47− 0.264y

Taking fuselage diameter of 3.79 m, the chord at y = 1.895 m is

cr(exposed) = 4.97 m

bexposedwing = 15.78− 3.79

2= 13.89 m

29

Swet = 2Sexposed

{1 + 1.2(t/c)avg

}(31)

Sexposedwing =1

2(4.97 + 1.31)× 13.89× 2 = 87.23 m2

Assuming (t/c)avg of 12.5%

Swet(exposedwing) = 2

(1 + 1.2(0.125)

)87.23 = 200.63 m2

Hence,

(CDo)W = 0.0025× 200.63

107.02= 0.004687

F1 = 1.52× 0.004687 = 0.007124

We also know that the drag polar is

CD = 0.0161 + 0.0482C2L

F2 =CDo − F1

W/S= 1.632× 10−6 m2/N

The above drag polar will not be valid at M greater than the Mcruise.Hence we need to estimate the drag polar (values of CDo and K) at Mmax.The drag divergence Mach number(MDD) for the aircraft is fixed atM = 0.82which is 0.02 greater than Mcruise. This would ensure that there is no wavedrag at Mcruise of 0.80. To estimate the increase in CDo from M = 0.80 toM = 0.84, we make a reasonable assumption that the slope of the CDo VsM curve remains constant in the region between M = 0.82 and M = 0.84.The value of this slope is 0.1 at M = 0.82. Hence, the increase in CDo isestimated as 0.02× 0.1 = 0.002.

From the data on B 787 available in website[2] we observe that the varia-tion in K is not significant in the range M = 0.82 to M = 0.84. Hence,valueof K is retained as in subcritical flow. However better estimates are used inperformance calculations presented later.

Consequently the drag polar that is valid at Mmax is estimated as

CD = 0.0181 + 0.0482C2L (32)

30

The change in the CDo is largely due to change in the zero lift drag of thewing, horizontal tail and vertical tail. This means that the change in CDoaffects F1 value alone.

Hence at Mmax

F1 = 0.009124

The value of F3 depends on the dynamic pressure at Vmax.

Vmax = Mmax ×√speed of sound at hcruise = 0.84× 295.2 = 248.1m/s

qmax =1

2ρV 2

max = 0.5× 0.364× 2482 = 11200.95

F3 =0.0482

11200.952= 3.84× 10−10 m4/N2

To obtain the optimum value of W/S, we minimize the thrust requiredfor Vmax. The relation between t(ie T/W ) and p is

tVmax = qmax

(F1

p+ F2 + F3p

)(33)

On minimizing tVmax , we get

poptimum =

√F1

F3

poptimum =

√0.009124

3.84× 10−10= 4873.31N/m2

The tVmax value at popt is found from eq.(33) as

tVmax = 0.06022

Allowing a 5 % extra thrust and using the new tVmax in eq. (33) gives twovalues of p viz.

p1 = 3344 Nm−2

p2 = 7101 Nm−2

Thus, any p between p1 and p2 would be acceptable from Vmax consider-ations with a maximum of 5% deviation from optimum.

3344 < p < 7101 N/m2

31

3.3 (R/C)max consideration

The value for (R/C)max at sea level was chosen as 700 m/min (11.67 m/s)which is typical for passenger airplanes.The thrust required for climb at cho-sen flight speed(V ) is related to (R/C) in the following way(section 4.2.4 oftext).

tR/c =R/C

V+q

pCD (34)

But, CD is

CD = F1 + F2p+ F3p2 (35)

q =1

2ρ0σV

2 (36)

∴ tR/C =R/C

V+

1

2ρ0σ

V 2

p(F1 + F2p+ F3p

2) (37)

The flight speed for optimum climb performance is not high and valuesof F1 and F2 correspond to their values for M < Mcruise. F3 is a function ofthe dynamic pressure.

Our motive is to find the minimum sea level static thrust (tsR/c) for various

values of V and then choose the minimum amongst the minima. For a givenV ,

popt =

√F1

F3

Therefore, a table is prepared for different values of velocity(Table 4)and the corresponding tR/C is obtained using eq.(37) and the correspondingvalue of F3. This tR/C is converted to tsR/C

by using the plots provided inReference 1.13, chapter 9. These plots provide the climb thrust variation forengine with bypass ratio 6.5 as a function of velocity and altitude. Usingthese plots,the tR/C is converted to tsR/C

.

32

V (m/s) popt tR/C tsR/C

80 1507 0.1893 0.2868100 2355 0.1637 0.2641120 3391 0.1487 0.2507140 4615 0.14 0.2469150 5298 0.1373 0.2483160 6028 0.1356 0.2510170 6805 0.1346 0.2554180 7629 0.1343 0.2617190 8500 0.1345 0.2691200 9419 0.1354 0.2780

Table 4: Variation of tR/C with p for (R/C)max

We observe that the value of tsR/Cremains low and almost constant for

a range of V values from 120 to 170 m/s. This provides a range of values ofp as given below

p1 = 3391 N/m2

p2 = 6793 N/m2

Therefore, for

3391 < p < 6805 N/m2

the climb performance is near the optimum.

3.4 Based on Minimum Fuel for Range (Wfmin)

In cruise flight, the weight of the fuel used (Wf ) is related to the range(R)and wing loading(p) as follows(section 4.2.5 of [5])

W f =R

3.6

√ρ0

2TSFC

√σq

(F1

p+ F2 + F3p

)(38)

The values of F1, F2, F3 corresponding to cruise conditions are as follows

F1 = 0.007124

F2 = 1.632× 10−6

33

Vcruise = Mcruise × 295.2 = 0.8× 295.2 = 236.3 m/s

qcruise = 0.5× ρ× V 2 = 0.5× 0.364× 236.32 = 10159.59 N/m2

F3 =0.0482

10159.59= 4.67× 10−10 m4/N2

Using eq.(38) we minimize W f and obtain poptimum as

poptimum =

√F1

F3

(39)

poptimum =

√0.007124

4.67× 10−10= 3905.84 N/m2

Using this value of p in eq.(38) along with R = 4000 km and TSFC =0.6hr−1, we get W fmin as

W fmin = 0.1514

Allowing an excess fuel of 5 % i.e. W fmin= 0.1590 and using eq.(38) we

get two values p1 and p2 as

p1 = 2676 N/m2

p2 = 5700 N/m2

Thus, any p within p1 and p2 would be acceptable from the point of viewof minimizing W f .

2676 < p < 5700N/m2

3.5 Based on Absolute Ceiling

At absolute ceiling, the flight is possible at only one speed. Observing thetrend of Hmax as hcruise + 0.6 km we choose the absolute ceiling to be Hmax

= 11.6 km. To find the tHmax , we solve the following two equations(section4.2.3 of [5]).

34

th =√

4K(F1 + F2p) (40)

th = 2qhmax

(F1

p+ F2

)(41)

The F1 and F2 values corresponding to this case are

F1 = 0.007124

F2 = 1.632× 10−6

In the absence of a prescribed velocity atHmax, the velocity correspondingto flight at (L/D)max is taken to calculate qmax. CL value corresponding toflight at (L/D)max is given by

CL =

√CDoK

=

√0.016

0.048= 0.577 (42)

qhmax =(W/S)

CL=

5500

0.577= 9532.06

The solution for popt is obtained by solving eqs.(40) and (41).

popt = 5500 Nm−2 as it should be.

thmax corresponding to poptimum is

thmax = 0.05581

Allowing a 5 % variation in Thrust, we get

thmax1 = 0.05302

thmax2 = 0.05860

The solutions to eq.(40) with the new thmax values are

p1 = 4567 Nm−2

p2 = 6547 Nm−2

Similarly, using in eq.(41), we get

35

p1 = 4942 Nm−2

p2 = 6201 Nm−2

From the above four values, the final lower and upper bounds from theceiling considerations are

p1 = 4942 Nm−2

p2 = 6201 Nm−2

4942 < p < 6201 N/m2

3.6 Summary of Constraints

We now tabulate the various constraints on the choice of W/S

Performance Criteria Allowable range of W/S in (Nm−2)sLand 3639 - 5328Vmax 3344 - 7101

(R/C)max 3391 - 6805

W f 2676 - 5700hmax 4942 - 6201

Table 5: Choice of (W/S)

From the table, we see that the allowable range of W/S values is

4942 < p < 5328 N/m2

3.7 Consideration of Wing Weight (Ww)

The weight of the wing depends on its area. According to Raymer[4], chapter15, for passenger airplanes, the weight of the wing is proportional to S0.649.Thus a wing with lower area will be lighter and for lower wing area, the W/Smust be higher. Hence we examine the advantage of choosing a higher wingloading than that indicated by minimum fuel requirement. It may be pointedout that the weight of wing structure is about 12% of Wg.

36

The optimum W/S from range consideration is 3906 N/m2 whereas witha 5% increase in Wf , the wing loading could go up to 5700 N/m2. If thewing loading of 5700 N/m2 is chosen, instead of 3906 N/m2, the weight ofthe wing would decrease by a factor of(

3906

5700

)0.649

= 0.782

Taking weight of the wing as 12% of Wg, the saving in the wing weightwill be 2.6%. However this higher wing loading will result in an increase inthe fuel by 5% of Wg. In the present case, Wf would be around 20% andhence 5% of Wf means an increase in the weight by 0.05× 0.2 = 1%.

Thus by increasing W/S from 3906 to 5700 N/m2, the saving in the Wg

would be around 2.6 - 1 = 1.6%. Thus it is advantageous to have higherW/S.

3.8 Choosing a W/S

We see from the Table 5 that a wide range of p is permissible which will stillsatisfy various requirement with permissible deviations from the optimum.To arrive at the final choice we consider the take-off requirement and choosehighest wing loading which would permit take-off within permissible distancewithout excessive (T/W) requirement. From data collection, the take-offdistance, balanced field length, is assumed to be 2150 m. From figure 5.4of Raymer(Reference 1.11) the take-off parameter {(W/S)/σCLt.o(T/W )} forthis field length is 180. With (W/S) in lb/ft2. We take σ = 1 (take-off at sealevel),CLt.o = 0.8× CLmax = 0.8× 2.5 = 2. Generally these types of aircrafthave (T/W) of 0.3.Substituting these values we get,

pfinal = 108.2 lb/ft2 = 5195 Nm−2

It is reassuring that this value of p lies within the permissible valuessummarized in Table 5.

3.9 Thrust Requirements

After selecting the W/S for the aircraft, the thrust needed for various designrequirements is obtained. These requirements decide the choice of engine.

37

3.9.1 Requirement for Vmax

We use the chosen value of p in the following equation

tVmax = qmax(F1

p+ F2 + F3p) (43)

and get the thrust required for Vmax at cruise altitude as(T

W

)Mmax

= 0.0602 (44)

Referring to engine charts in Jenkinson[8], chapter 9, for a turbo fanengine with bypass ratio of 6.5, the sea level static thrust is

T

W=

0.0602

0.18= 0.334 (45)

In our case, this would mean a Thrust requirement of

Treq = 193.9 kN

3.10 Requirements for (R/C)max

As in the case for Vmax, we use our final design choice for (W/S) in thefollowing equation,

tR/c =R/C

V+

1

2ρ0σ

V 2

p(F1 + F2p+ F3p

2) (46)

Substituting appropriate values, we get(T

W

)R/C

= 0.252 (47)

In our case, this would mean a thrust requirement of

Treq = 146.3kN

3.11 Take-Off Thrust Requirements

The take of (T/W ) is taken to be 0.3(choice is motivated by similar aircraft).This implies a thrust requirement of

Tto = 0.3 ∗Wg = 174.2 kN

38

3.12 Engine Choice

From the previous section, we see that the max. Thrust requirements occursfrom Take off considerations.

Tmax = 193.9 kN

As we have adopted a twin engine design, this means a per engine thrustof

Tmax = 96.95 kN/engine

We look for an engine which supplies this thrust and has a TSFC of0.6hr−1 and bypass ratio of around 6.5. Some of the engines with perfor-mance close to these numbers are taken from Jenkinson[8], chapter 9 andwebsite[1].

Finally, we chose CFM56-2B model of turbofan with a sea level staticthrust of 97.9 kN as this engine satisfies nearly all our requirements.

3.13 Engine Characteristics

For performance analysis, the variation of thrust and TSFC with speed andaltitude are required. Jenkinson[8], chapter 9 has given non dimensionalcharts for turbo fan engines with different bypass ratios. Choosing the chartsfor bypass ratio = 6.5 and sea level static thrust of 97.9kN , the engine curvesare presented below.

39

Fig

ure

5:C

ruis

eT

hru

stper

engi

ne

for

variou

sal

titu

des

40

Fig

ure

6:Var

iati

onof

Clim

bT

hru

stw

ith

Alt

itude

and

Mac

hN

o.(B

ypas

sra

tio

=6.

5)

41

4 Wing Design

4.1 Introduction

The weight and the wing loading of the airplane have been obtained in sec-tions 2 and 3 as 59175 kgf(579915 N) and 5195 N/m2 . These give wing areaas 111.63 m2 . The wing design involves choosing the following parameters.

1. Airfoil selection

2. Aspect ratio

3. Sweep

4. Taper ratio

5. Twist

6. Incidence

7. Dihedral

8. Vertical location

In the following subsections, the factors affecting the choice of parametersare mentioned and then the choices are effected.

4.2 Airfoil Selection

The airfoil shape influences CLmax , CDmin, CLopt , Cmac and stall pattern.

These in turn influence stalling speed, fuel consumption during cruise, turn-ing performance and weight of the airplane.

For high subsonic airplanes, the drag divergence Mach number(MDD) isan important consideration. It may be recalled that (MDD) is the Machnumber at which the increase in the drag coefficient is 0.002 above the valueat low subsonic Mach numbers. A supercritical airfoil is designed to increaseMDD. NASA has carried out tests on several supercritical airfoils and recom-mends the use of NASA-SC(2) series airfoil with appropriate thickness ratioand camber.

42

4.2.1 Design Lift Coefficient

The airfoil will have a Clopt at which it’s drag coefficient is minimum. Forgeneral design the airfoil is chosen in such a way that the CLcruise

of theairplane is equal to the Clopt of the airfoil.

CLcruise=

(W/S)

qcruise(48)

Using the value of (W/S) = 5195 Nm−2 and the q corresponding toM = 0.8 at 11 km altitude, we get

Clcruise= 0.512 (49)

For choice of thickness ratio and wing sweep, we take Clopt = 0.5.

4.2.2 Airfoil Thickness Ratio and Wing Sweep

Airfoil thickness ratio(t/c) has a direct influence on drag, maximum lift, stallcharacteristics, structural weight and critical Mach number. A higher t/c im-plies a lower critical Mach number but also a lower wing weight.Thus we needto choose an optimum t/c for the airfoil.

Clopt = 0.5 has been chosen and the cruise Mach number is 0.8. In orderto ensure that the drag divergence Mach number is greater than Mcruise, wechoose MDD as 0.82. This is based on the consideration that there shouldbe no increase in drag at Mcruise, ∆CDwave is 0.002 at MDD and the slope ofthe CD Vs M curve around MDD is 0.1 . NASA[3] gives experimental resultsfor several super-critical airfoils with different (t/c) and Clopt . Curves forClopt = 0.4, 0.7, 1.0 are available in the aforesaid report. We interpolate andobtain the curve for Clopt = 0.5.

The MDD for the wing can be estimated in the following manner.

MDD = (MDD)a/f + ∆MA + ∆MΛ (50)

where ∆MA and ∆MΛ are corrections for influences of the aspect ratioand the sweep.

The change in MDD with A is almost zero for A > 8. Since we havechosen A = 9.3, the second term in the above equation will not contribute toMDD. Further from Hoerner[9], chapter 15, the change in MDD due to sweepis given as

1− Λ

90=

1−MDDΛ

1−MΛ=0

(51)

43

The supercritical airfoil with (t/c) = 14% has MDD = 0.74 at CLopt of0.5. Using this in eq.(51) we obtain Λ which would give MDDΛ of 0.82,

1− Λ

90=

1− 0.82

1− 0.74

∴ Λ = 27.7◦

The average thickness has been arrived at as 14 %. However, to reducestructural weight, the (t/c)root is increased and the (t/c)tip is decreased, Con-sidering the features for Airbus A310 and Boeing B 767 which haveMcr = 0.8 and similar values of Λc/4, it is decided that the variation of (t/c)along span be such that (t/c) of 15.2% at root, (t/c) of 11.8% at spanwiselocation of the thickness break and (t/c) of 10.3% at the tip.

Thickness break location is the spanwise location upto which the trailingedge is straight. From the data collection this location is at 34% of semispan.

4.3 Other Parameters

4.3.1 Aspect Ratio

The aspect ratio affects CLα , CDiand wing weight. The value of CLα de-

creases as A decreases. For example, in the case of an elliptic wing,

CLα =A

A+ 2(Clα)a/f (52)

The induced drag coefficient can be expressed as

CDi=C2L

πA(1 + δ) (53)

where δ depends on A, λ and Λ. A high A increases the span of the wingwhich in turn requires more space in the hangar. A higher Aspect ratio wouldalso result in poor riding quality in turbulent weather. All these factors needcareful optimization. However at the present stage of design we chooseA = 9.3 based on trends indicated by data collection.

Correspondingly, the wing span would be

b =√AS = 32.22m

44

4.3.2 Taper Ratio

Wing taper ratio is defined as the ratio between the tip chord and the cen-terline root chord. Taper ratio affects the

• Induced drag

• Weight

• Tip stalling

Induced drag is low for taper ratios between 0.3-0.5. Lower the taper ratio,lower is the weight. A swept wing also has higher structural weight thanunswept wing. Since the present airplane has a swept wing, a taper ratio of0.24 has been chosen based on the trends of current swept wing airplanes.

4.3.3 Root and Tip Chords

Root chord and tip chord of the equivalent trapezoidal wing can now beevaluated.

cr =2S

b(1 + λ)= 5.59 m

ct = crλ = 1.34 m

c =2

3

(1 + λ+ λ2)

(1 + λ)cr = 3.9 m

Location of the quarter chord of the mac from wing leading edge at theroot is 4.76 m

4.3.4 Dihedral

The Dihedral is the angle of the wing with respect to the horizontal whenseen in the front view .Dihedral of the wing affects the lateral stability of theairplane.Since there is no simple technique for arriving at the dihedral anglethat takes all the considerations into effect we need to initially choose a di-hedral angle based on data collected(Table A). Hence we choose a reasonablevalue for the dihedral as

Γ = 5o

4.3.5 Wing Twist

We have assumed a linear twist of 3o.

45

4.4 Cranked Wing Design

If we observe the design of current high subsonic airplanes, we see that thetrailing edge is ’straight’ for a part of the span, in the inboard region. Alarger chord in the inboard region has the following advantages

1. more space for fuel and landing gear

2. the lift distribution is changed such that more lift is produced in theinboard section which reduce the bending moment in the root.

This type of design is called a wing with cranked trailing edge. The valueof the span upto which the trailing edge is straight has to be obtained byoptimization. However at the present stage of design, based on the currenttrends, the trailing edge is made unswept till 35% of semi span. Root chordof the cranked wing is

crcranked = 7.44 m

Span of wing portion with unswept trailing edge = 0.35× 32.22 = 11.28 m

Figure 7: Plan View of Cranked Wing

46

4.5 Wing Incidence(iw)

The wing incidence angle is the angle between wing reference chord andfuselage reference line. Wing incidence angle is chosen to minimize drag atsome operating conditions,usually cruise.The incidence angle is chosen suchthat when the wing is at the correct angle of attack for the selected designcondition,the fuselage is at the angle of attack for minimum drag(usually atzero angle of attack). Usually wing incidence is ultimately set using windtunnel data.However, for an initial estimate for our preliminary design weproceed as follows

CLcruise= CLα(iw − α0L) (54)

In the present case,CLcruise

= 0.512

CLα is computed using the following formula in Raymer[4], chapter 12,

CLα =2πA

2 +√

4 + A2β2

η2 (1 + tan2Λmax

β2 )(SexpSref

)(F ) (55)

where,

β2 = 1−M2

η = 1

F = 1.07

(1 +

d

b

)2

Sexp = area of exposed wing

Substituting various values, we get

CLα = 6.276 rad−1

αL=0 for the airfoil was calculated using camber line of the supercriticalairfoil with 14% thickness ratio. The value is −5.8◦. Substituting the valuesyields a value of iw which is negative. This can be attributed to the fact thatthe airplane is flexible. Hence the value of iw is chosen from similar airplanes.

iw = 1◦

which is the value recommended in Raymer[4], chapter 4.

4.6 Vertical Location of Wing

The wing vertical location for the designed airplane has been chosen to be alow wing configuration which is typical of similar airplanes.

47

4.7 Areas of Flaps and Ailerons

These areas are chosen based on the initial data collection of similar aircraft.

1. Trailing edge : Fowler flaps.

2. Leading edge : full span slats.

We choose

SflapS

= 0.17

SslatS

= 0.1

SailS

= 0.03

5 Fuselage and Tail Layout

5.1 Introduction

The fuselage layout is important in the design process as the length of theairplane depends on this.The length and diameter of the fuselage are relatedto the seating arrangement.

The Fuselage of a passenger airplane can be divided into four basic sec-tions viz. nose, cockpit, payload compartment and tail fuselage. In thissection, the fuselage design is carried out by choosing the various parame-ters.

5.2 Initial Estimate of Fuselage Length

By observing the lf/b of similar airplanes, we get the first estimate of lffor the present case. The lf/b value chosen is 1.05. Using b = 32.22m asobtained from wing design, the Fuselage length is 33.83 m.Raymer[4], chapter 6 provides a relation between gross weight and length offuselage as follows.

lf = aW co (56)

48

where Wo is in lbs and lf in ft. For a jet transport airplane, a = 0.67and c = 0.43. Using Wo = 59175 × 2.205 lbf , an lf of 31.83 m is obtained.This is in good agreement of the value obtained based on data collection.

5.3 Nose and Cockpit - Front Fuselage

The front fuselage accommodates the forward looking radar in the nose sec-tion, the flight deck with associated windscreen, and the nose undercarriage.Anthropometric data for flight crews has provided the basis for the arrange-ment of pilot’s seats, instruments and controls. Development of electronicdisplays has transformed the traditional layout of the flight deck. The air-craft must be capable of being flown from either pilot seat position; thereforethe wind screen and front geometry will be symmetrical about the aircraftlongitudinal center line. Modern ’glass’ cockpit displays and side stick con-trollers have transformed the layout of the flight deck from the traditionalaircraft configuration. The front fuselage profile presents a classical designcompromise between a smooth shape for low drag and the need to have flatsloping windows to give good visibility. The layout of the flight deck andthe specified pilot window geometry is often the starting point of the overallfuselage layout.

For the current design, the flight deck of various similar airplanes areconsidered and the following value of lnose/lf and is chosen.

lnoself

= 0.03

For the cockpit length (lcockpit), standards have been prescribed by Raymer(Reference 1.11,chapter 9). lcockpit for the two member crew is chosen as 100inches (2.5 m).

lcockpit = 2.5 m

5.4 Passenger Cabin Layout

Two major geometrical parameters that specify the passenger cabin areCabin Diameter and Cabin Length. These are in turn decided by morespecific details like number of seats, seat width, seating arrangement (num-ber abreast), seat pitch, aisle width and number of aisles.

49

5.4.1 Cabin Cross Section

The shape of the fuselage cross section is dictated by the structural require-ments for pressurization. A circular shell reacts the internal pressure loadsby hoop tension. This makes the circular section efficient and therefore low-est in structural weight. However a fully circular section may result in toomuch unusable volume above or below the cabin space. This problem isovercome by the use of several interconnecting circular sections to form thecross-sectional layout. The parameters for the currently designed airplaneare arrived at by considering similar airplanes(Table A).

We choose a circular cross section for the fuselage.

The overall size must be kept small to reduce aircraft weight and drag,yetthe resulting shape must provide a comfortable and flexible cabin interiorwhich will appeal to the customer airlines. The main decision to be taken isthe number of seats abreast and the aisle arrangement.The number of seatsacross will fix the number of rows in the cabin and thereby the fuselagelength.Design of the cabin cross section is further complicated by the need toprovide different classes like first class, business class, economy class etc.

5.4.2 Cabin length

Following the trend displayed by current aircraft, we choose to have twoclasses viz Economy class and Business class.The total number of seats(150)is distributed as 138 seats in the economy class and 12 seats in the businessclass.Cabin parameters are chosen based on standards for similar airplanes. Thevarious parameters chosen are as follows

Parameter Economy Class Business classSeat Pitch (in inches) 32 38Seat width (in inches) 20 22Aisle width (in inches) 22 24Seats abreast 6 4Number of Aisles 1 1Max. Height (in m) 2.2 2.2

Since the business class has a 4 abreast seating arrangement,the numberof rows required will be 3 and the economy class will have 23 rows.The cabinlength is found out by using the seat pitch for each of the classes.

50

Class No.of seats No.of rows Seat Pitch (in) Cabin length(m)Economy 138 23 32 18.4Business 12 3 38 2.85

Hence,the total cabin length will be 18.4 + 2.85 = 21.25 m.

5.4.3 Cabin Diameter

Using the number of seats abreast,seat width,aisle width we calculate theinternal diameter of the cabin.

df(internal) = 22× 1 + 19× 6 = 136 in = 3.4 m

According to the standards prescribed by Raymer[4], chapter 9, the struc-tural thickness is given by

t = 0.02df + 1′′ = 0.02× 136 + 1 = 3.72 in = 0.093 m

Therefore the external diameter of the fuselage is obtained as 3.4+0.093×2 = 3.59 m.

5.5 Rear Fuselage

The rear fuselage profile is chosen to provide a smooth, low drag shape whichsupports the tail surfaces. The lower side of the profile must provide ade-quate clearance for aircraft when rotation during take off. The rear fuselageshould also house the auxillary power unit(APU).

Based on data collected for similar aircraft we choose the ratio ltail/lf as0.25.

5.6 Total Fuselage Length

The cabin length and cockpit length have been decided to be 32.08 m and3.3 m respectively.We have also chosen the ratios of nose and tail length withrespect to lf as 3% and 25%. Thus cabin and cockpit length form 72% of lf .Hence the fuselage length is calculated as 23.75/0.72 = 33 m.The lengths of

51

various parts of the fuselage are indicated below

Nose length = 1 m

Cockpit length = 2.5 m

Cabin length = 21.25 m

Rear length = 8.25 m

Total = 33 m

5.7 Tail surfaces

The type and area of the tail surfaces are important in determining the sta-bility of the airplane. A conventional tail arrangement is chosen. Some ofthe important parameters that decide the aerodynamic characteristics of thetail are area ratio (St/S), tail volume ratio(VH and VV ), tail arm, tail spanetc. All these parameters have to be decided for both the horizontal andvertical tail.

From data collected on similar airplanes, we choose the following valuesfor the tail parameters.

Parameter Horizontal Tail Vertical TailArea ratio (St/S) 0.31 0.21Aspect ratio 5 1.7Taper ratio 0.26 0.31

• Area

The Areas of the horizontal and vertical tail(Sh and Sv) are calculatedas

Sh = 0.31× 111.63 = 34.61 m2

Sv = 0.21× 111.63 = 23.44 m2

• Span

The span of the horizontal and vertical tail (bh and bv) are given as

bh =√AhSh (57)

bv =√AvSv (58)

52

Taking ARH = 5 and ARV = 1.7, we get

bH = 13.15 m

bV = 6.31 m

• Root and tip chords

The chord lengths of the horizontal and vertical tail are obtained as

crh =2Sh

bh(1 + λh)= 4.18 m

crv =2Sv

bv(1 + λv)= 5.67 m

cth = λcrh = 1.09 m

ctv = λcrv = 1.76 m

• Tail arm

Tail arm is the distance between the wing aerodynamics center andthe tail(horizontal or vertical) aerodynamic center.The value of the tailarm is chosen based on data collection. ratio.

Choosing lh as 45% of lf and lv as 42% of lf yield,

lh = 14.85 m

lv = 13.86 m

VH =ShlhSc

(59)

VV =SvlvSb

(60)

Hence,

VH = 1.18

VV = 0.09

53

5.8 Engine Location

The type of Engine mounting and it’s location play a major role in decidingthe overall drag coefficient of the airplane. A conventional wing mounted en-gine is chosen as it facilitates periodic maintenance in an industry where anunscheduled downtime could mean huge losses to the airliners. The enginesare attached to the lower side of the wing using pylons to reduce drag. Theother reason for choosing a wing mounted engine is the fuel is stored in thewings itself, thereby reducing the length of the fuel line.

From the data collection of similar airplanes, the engine location is fixedat 34% of the semi span.

5.9 Landing Gear Arrangement

One of the principal moving parts on the aircraft is the landing gear. Thismust be light, small, provide good ride dynamics during taxiing and safe en-ergy absorption at touch down. It must be retractable to reduce drag duringflight. So housing of the landing gear is a space constraint.A conventional tri-cycle landing gear is chosen based on trend followed by similar aircraft. Theimportant parameters of this type of landing gear are wheel track, wheel baseand turning radius. The values of the parameters(shown below)were basedon data collected from similar aircraft.

Parameter ValueWheel base (in m) 13.2Track length (in m) 5.8

Turning radius (in m) 19.3

54

6 Estimation of Component Weights and C.G

Location

Aircraft weight is a common factor which links different design activities(aerodynamics, structures, propulsion, layout, airworthiness, environmental,economic and operational aspects).To this end, at each stage of the design,acheck is made on the expected total mass of the completed aircraft. A sepa-rate design organization(weights department)is employed to assess and con-trol weight.In the preliminary design stage,estimates have to made from his-torical statistical data of all the component parts of the aircraft from similarairplanes. As parts are manufactured and the aircraft prototype reaches com-pletion it is possible to check the accuracy of the estimates by weighing eachcomponent and where necessary instigate weight reduction programmes.

6.1 Aircraft mass statement

The weight of the entire airplane can be sub-divided into empty weight anduseful load. The empty weight can be further subdivided into

• Structures group

• Propulsion group

• Equipment group

DCPR(Defense Contractor Planning Report) weight is taken as the weightobtained after deducting weights of wheels, brakes, tires, engines, starters,batteries, equipments, avionics etc from the empty weight.DCPR weight isimportant for cost estimation, and can be viewed as the weight of the parts ofthe airplane that the manufacturer makes as opposed those of items boughtand installed.

It has become normal practice in aircraft design to list the various com-ponents of aircraft mass in a standard format.The components are groupedin convenient subsections as shown below.

6.1.1 Structures Group

1. Wing(including control surfaces)

2. Tail(horizontal and vertical including controls)

3. Body(or fuselage)

55

4. Nacelles

5. Landing gear (main and nose units)

6. Surface controls

6.1.2 Propulsion Group

1. Engine(s)(dry weight)

2. Accessory gearbox and drives

3. Induction system

4. Exhaust system

5. Oil system and cooler

6. Fuel system

7. Engine controls

8. Starting system

9. Thrust reversers

6.1.3 Fixed equipment group

1. Auxiliary power unit

2. Flight control systems(sometimes included in structural group)

3. Instruments and navigation equipment

4. Hydraulic systems

5. Electrical systems

6. Avionics systems

7. Furnishing

8. Air conditioning and anti-icing

9. Oxygen system

10. Miscellaneous(e.g.fire protection and safety systems)

56

6.2 Weights of Various Components

After making the classification between various groups and listing the com-ponents in each group,we next proceed to determine the weights of thesecomponents.

In the preliminary design stages it is not possible to know the size of indi-vidual aircraft components in great detail but it is possible to use predictionmethods that progressively become more accurate as the aircraft geometryis developed.Most aircraft design textbooks contain a set of equations empir-ically derived based on existing aircraft. For the present design, we chooseto follow equations prescribed in Appendix 8.1 of [5]. Using these equations,the weights of various individual components are calculated.

6.3 C.G Location and C.G Travel

6.3.1 Wing Location on Fuselage

The wing longitudinal location is decided based on the consideration the C.Gof the entire airplane with full payload and fuel is around the quarter chordof the m.a.c.We tabulate the weights and the corresponding C.G locationsof various components and then apply moment equilibrium about the noseof the airplane in order to solve for Xl.e(the distance of leading edge of rootchord of the wing from the nose).In tabulating the results,we assume thatthe C.G locations of wing, horizontal tail and vertical tail are at 40% of therespective m.a.c.The fuselage C.G is taken to be at 42% of it’s length.Theengine C.G location was taken to be at 40% of it’s length.The distance ofthe engine C.G from the root chord was measured for various airplanes andwe chose a distance of 2 m.All other components were taken to have a netC.G location at 42% of the fuselage length.The tabulated values are givenbelow.The nose wheel was placed at 14% of the fuselage length and the mainlanding gear position was determined by using the wheelbase from section 5.

Remark

• Using data for equivalent trapezoidal wing in section 4, the locationof wing c.g. is at 5.34 m behind the leading edge of the root chord.The quarter chord of m.a.c is at 4.76 m behind the leading edge of rootchord.

• Noting that the tail arm is 14.85 m and that the c.g of tail is 15 %behind the a.c., the distance of horizontal tail c.g. from leading edge

57

of root chord of wing is 20.05 m. In a similar way, c.g. of vertical tailis at 19.56 m behind leading edge of the root chord of wing.

Component Weight(kg) C.G Location from Nose(m)Wing 5855.41 Xle+5.34

Fuselage 6606.60 13.86Horizontal tail 1160.94 Xle+20.05Vertical tail 746.22 Xle+19.56

Engine group 5659.19 Xle+2Nose Wheel 363.18 4.62

Main landing gear 1961.25 17.82Fixed equipment total 7421.09 13.86

Fuel 12130.88 Xle+4.76Payload 17270 14.13

Gross Weight 59175 Xle+4.76

By applying moment equilibrium about the nose of the airplane,we obtainlocation of wing leading edge at the root to be 9.85 m from the nose of theairplane.

The C.G of the airplane lies at 14.61 m from the nose.

6.4 C.G Travel for Critical Cases

6.4.1 Full Payload and No Fuel

For the case of full payload and no fuel,the fuel contribution to the weightis not present.However, since we have assumed that the c.g of the fuel tobe at the quarter chord of the m.a.c of the wing(where the c.g of the entireairplane has been positioned)there will be no c.g shift in this case.

Hence,the C.G shift is 0%.

6.4.2 No Payload and No Fuel

For this case,the fuel as well as the payload contribution are not present.Sincethe c.g of payload is not at the c.g of the entire airplane,the c.g is bound toshift by a certain amount in this case.The moment calculations were per-formed and the new c.g location was obtained as 14.93 m.Therefore the c.gshift is 14.93 - 14.63 = 0.3 m.

Hence the c.g shift is +7.28% of m.a.c.

58

6.4.3 No Payload and Full fuel

For this case,since there is no payload, the c.g is bound to shift.On perform-ing calculations,we obtain the new c.g location to be 14.84 m.Therefore thec.g shift is 14.84 - 14.63 = 0.21 m.

Hence the c.g shift is +5.17% of the m.a.c.

6.4.4 Payload distribution for 15% c.g travel

According to Lebedenski[7], a total c.g shift of 15% is acceptable in generalfor commercial airplanes.Hence,we go on to obtain the maximum payloadthat can be concentrated in the front portion of the passenger cabin suchthat a c.g shift of 7.5% is produced.

We assume the percentage of payload to be x and also assume the pay-load c.g to be at x% of the passenger cabin length.Performing the calculationsyields the value of x to be 90%.

Similarly,we also obtain the maximum payload that can be concentratedat the rear half of the passenger cabin resulting in a c.g shift of 15%.

On performing calculations we obtain the value of x as 70%.

Hence,the c.g locations for various critical cases and payload distributionfor c.g shift of 15% have been calculated.

6.5 Summary

• Wing location(leading edge of root of trapezoidal wing) - 9.85 m

• c.g location with Full payload and full fuel - 14.61 m

• c.g travel for No Payload and No Fuel - 7.28%

• c.g travel for No Payload and full Fuel - 5.17%

• For a c.g travel of 7.5% on either side of original c.g location,90% ofpassengers can be concentrated in the front and 70% in the rear.

59

7 Control Surfaces

7.1 Stability and Controllability

The ability of a vehicle to maintain its equilibrium is termed stability andthe influence which the pilot or control system can exert on the equilibriumis termed its controllability.The basic requirement for static longitudinal sta-bility of any airplane is a negative slope of the curve of the pitching momentcoefficient, Cmcg, versus lift coefficient,CL. Dynamic stability requires thatthe vehicle be not only statically stable,but also that the motions followinga disturbance from equilibrium be such as to restore the equilibrium.

Even though the vehicle might be statically stable, it is possible that theoscillations following a disturbance might increase in magnitude with eachoscillation,thereby making it impossible to restore the equilibrium.

7.2 Static Longitudinal Stability and Control

7.2.1 Specifications

• The horizontal tail must be large enough to insure that the static longi-tudinal stability criterion,dCmcg/dCL will be negative for all anticipatedcenter of gravity positions.

• An elevator should be provided so that the pilot will be able to trimthe airplane(maintain Cm = 0) at all anticipated values of CL.

• The tail should be large enough and and its elevator powerful enoughto enable the pilot rotate the airplane during the take-off run to therequired angle of attack.This condition is termed as the Nose wheelLift-off condition.

7.2.2 Aft Center of gravity limit

For the “stick free” case and for small angles of attack,the following expres-sion for the aft center of gravity limit in terms of the tail-size parameter,Vwe have the following equation. (Section 9.2 of [5])

(xc.g)aft = xa.c−(dCmdCL

)Fus,Nac

+atawV ηt

(1− dε

dα

)(1−Chατ

Chδ

)+

(dCmdCl

)power

(61)The value of xc.g from above equation is termed the “stick-free neutral

point”,since it is the c.g location at which the static stability is neutral.

60

7.2.3 Forward center of Gravity Limit

The forward c.g. limit is not generally dependent on maintaining stability.As the c.g is moved forward ,the stability contribution xc.g−xa.c of the wingbecomes more and more negative ,thereby increasing the static stability.Inorder to keep the airplane in equilibrium as the c.g is moved forward,theelevator must be capable of trimming out the resulting negative pitchingmoment.The pitching moment will be the greatest when the airplane is atCLmax when the airplane is landing and ground effects decrease the down-wash at the tail.

The equation of pitching moments may be solved for the position of themost forward c.g by assuming the airplane trimmed(Cmcg = 0) at CLmax asfollows(Section 9.2 of [5])

(xcg)forward = xac−Cmδ

CLmax

[δemax+

αw − εG − iw + itτ

+Cmac(flaps) + Cm(fus) + Cm(power)

Cmδ

](62)

7.2.4 Determination of initial parameters

• (dCmdCL

)Fus (dCmdCL

)Fus

=KfW

2fLf

Scaw(63)

The value ofKf is obtained as 0.0119 from graph 1-9:1 of K.D.Wood[10].

aw=6.276 /radian = 0.1095 /degree

from the value obtained in section 4.5 on wing design.

Therefore, (dCmdCL

)fus

=0.0119× 3.592 × 33

111.63× 3.9× 0.1095= 0.1036

The contribution of nacelle to (dCm/dCL) is neglected.

• dε/dα

dε

dα=

114.6× awπA

(64)

61

dε

dα=

114.6× 0.1095

π × 9.3= 0.4297

•(dCm

dCL

)power (

dCmdCL

)power

=TtpWc

(65)

tp is the distance of thrust line from c.g(the distance is measured per-pendicular to the thrust line).For the designed airplane we make anestimate of tp to be 0.19 m.At the cruise altitude, we choose a (T/W )of 0.06.

Therefore, (dCmdCL

)power,cruise

=0.06× 0.62

13= 0.00292

• (CL)max is taken as 2.5 from Section 3. (CL)max with no flaps is 1.4.(∆CL)flaps = 1.1.

• awg is the lift curve slope of the wing close to the ground. It is ob-tained by calculating the value of aw at lower velocities. A value ofV = 1.3× 49 = 63.7m/s corresponds to a value of M = 0.19 and hencegives a value of

(aw)landing = 4.57/radian = 0.0796.

The awg is obtained by adding the ground effect to the (aw)landing ob-tained.Hence

(awg)landing = 1.1(aw)landing = 5.027/radian = 0.0877/deg (66)

• αWg

αWg =(CL)maxawgk

(67)

k is the ground effect factor obtained from Fig 1-9:4 of Wood[10].(CL)max is the value without flaps and corresponds to 1.4. k was ob-tained as 1.1((for height of a.c above ground)/semi span of 0.1).

αWg = 10.16◦

62

• at and atg

at is obtained as 0.0828/deg by using the tail parameters in eq.(55).atg is the lift curve slope of the wing close to the ground. It is ob-tained by calculating the value of at at lower velocities. A value ofV = 1.3× 49 = 63.7m/s corresponds to a value of M = 0.19 and hencegives a value of

(at)landing = 3.91/radian = 0.0682/deg.

The atg is obtained by adding the ground effect to the (at)landing ob-tained.Hence

(atg)landing = 1.1(at)landing = 5.027/radian = 0.0877/deg (68)

• iw is taken as 1◦ from Section 4.

• Cmjet at landing = 0

• Cmac(flaps)

Cmac(flaps) = Cmac + ∆Cmac(f)SfcfSc

(69)

Cmac for the airfoil is taken as −0.1 from airfoil database.∆Cmac istaken as -0.4 from Perkins and Hage[11], Figure 5.40.

Cmac(flaps) = −0.1− 0.4 ∗ 0.56 ∗ 1.1 = −0.3464

• Cm(Fus) (dCmdα

)fus

=

(dCmdCL

)fus

CLalpha(70)

Hence using the value of CLα with ground effect,

(Cmα)fus = 0.1036× 0.0877 = 0.0091

Cmfus= 0.0091× (αw − iw) = 0.0091(10.16− 1) = 0.0834

63

• Chα and Chδ

The values of Chα and Chδ are obtained from Fig 1-9:5 of Wood[10].Since not much detail is available about the nature of elevators weassume the standard design and obtain the following values.

Chα=-0.00660Chδ=-0.01140

• Cmδ

Cmδ = −atStS

ltcηtτ (71)

Cmδ = −0.08095× 0.95× 0.57× VH = −0.04438VH

• δemax

δemax is chosen as −25◦ which is typical of most airplanes.

• it For the preliminary design we assume it = 1 which is the typicallythe value of passenger airplanes.

Now,that we have obtained the various parameters required for the longi-tudinal stability criterion we go on to calculate V which affects the horizontaltail sizing. We adopt the following consideration to determine V . Cmα isapproximately equal to -1.15 for transport airplane at M = 0.8(Raymer[4],chapter 16). Assuming c.g at a.c(

dCmdCL

)=−1.15

6.276= −0.183

Hence

xcg(aft)

c− xac

c= −0.183

Substituting in eq.(61), we get

−0.183 = 0.1036− 0.2958V + 0.00292

∴ V = 0.98

We obtain the horizontal tail area to be

64

Sht =0.98× 3.9× 111.63

14.86= 28.71m2

Remark: Keeping in view the large number of approximations involved incalculation of parameters during landing and take-off, the cross check forforward c.g. location and nose wheel lift-off conditions are not carried out atthis stage.

7.3 Lateral Stability and Control

7.3.1 Specifications

• The directional stability criterion,dCn/dCψ should be negative for anyanticipated speed greater than 1.2 times the stalling speed.

• The yawing moment control(rudder) must be powerful enough to (a)counteract the yawing moment encountered in rolls(”adverse yaw”),(b)in cross-wind landings or takeoffs, (c)in one engine off conditionand (d)in spin when the recovery is effected primarily by the ruddercontrol.

• To regain and maintain straight flight with one engine inoperative at aminimum speed not greater than 1.3 times the stalling speed.

7.3.2 Equations for directional stability

The equations for directional stability can be derived as

dCndψ

= Cnψ(wing) + Cnψ(Fus) + Cnψ(power) + Cnψ(Tail) (72)

7.3.3 Determination of initial parameters

In the preliminary analysis of directional static stability, the contributions ofwing, power and interference effects are ignored.

• Cnψ(fus)

Cnψ(fus) =knVn

28.7Sb(73)

The value of kn was obtained from Figure 1:9-2 of Wood[10] as kn=0.95

Cnψ(fus) =0.95× 217.86

28.7× 111.63× 32.22= 0.002005

65

• Cnψ(tail)

Cnψ(tail) = −avSvS

lvb

(74)

av = 0.0378 per degree.

Cnψ(tail) = −0.0378× VV

The value of Cnψ(desirable) is given by Perkins and Hage[11] as follows

Cnψ(desirable) = −0.0005

[W

b2

]1/2

(75)

Therefore ,for the present case we have,

Cnψ(desirable) = −0.001709

Hence,

Cnψ(desirable) = Cnψ(fus) + Cnψ(tail) (76)

−0.001709 = 0.002005− 0.0378× VV (77)

or VV = 0.098This value is almost the same as what we obtained in our initial tail

sizing.Therefore,vertical tail area is

Svt =111.63× 32.22× 0.098

13.86= 25.43m2

66

8 Features of the Designed Airplane

8.1 Three View Drawing

The 3-view drawing of the airplane designed is given in figure

8.2 Overall Dimensions

Length : 34.32Wing Span : 32.22 mHeight above ground : 11.17Wheel base : 13.2 mWheel track : 5.8 m

8.3 Engine details

Similar to CFM 56 - 2BSeal Level Static Thrust : 97.9 kNBy pass ratio : 6.5 (For which the Engine characteristics are given in [8] )SFC : at M = 0.8, h = 10972 m(36 000 ft), SFC is taken as 0.6 hr−1

8.4 Weights

Gross Weight : 59175 kgfEmpty Weight : 29706 kgfFuel Weight : 12131 kgfPayload : 17338 kgfMaximum Landing Weight : 50296 kgf

67

Fig

ure

8:T

hre

evie

wdra

win

gof

the

airp

lane

68

8.5 Wing Geometry

Planform Shape : Cranked wingSpan : 32.22 mArea : 111.63 m2

Airfoil : NASA - SC(2) series, t/c = 14%, Clopt = 0.5Root Chord : 5.59 m (Equivalent Trapezoidal wing)Tip Chord : 1.34 m (Equivalent Trapezoidal wing)Root Chord of Cranked Wing : 7.44 mPortion of wing with straight trailing edge : 11.28 mMean Aerodynamic Chord : 3.9 mQuarter chord Sweep : 27.69o

Dihedral : 5o

Twist : 3o

Incidence : 1.4o

Taper Ratio : 0.24 (Equivalent Trapezoidal wing)Aspect Ratio : 9.3

8.6 Fuselage Geometry

Length : 33 mMaximum Diameter : 3.59 m

8.7 Nacelle Geometry

No. of nacelles : 2Nacelle Diameter : 1.62 mCross sectional Area : 2.06 m2

Length of Nacelle : 3.3 m (based on B737 Nacelle)

8.8 Horizontal Tail Geometry

Span : 11.98 mArea : 28.71 m2

Mean Aerodynamic Chord : 2.67 mQuarter Chord Sweep : 32o

Root Chord : 3.80 mTip chord : 0.99 m

69

Taper Ratio : 0.26Aspect Ratio : 5

8.9 Vertical Tail Geometry

Span : 6.58 mArea : 25.43 m2

Root Chord : 5.90 mTip chord : 1.83 mMean Aerodynamic Chord : 4.22 mQuarter Chord Sweep : 37o

Taper Ratio : 0.31Aspect Ratio : 1.70

8.10 Other details

CLmax without flap : 1.4CLmax with landing flaps : 2.7Maximum Load Factor nmax : 3.5CLmax with T.O flaps : 2.16

8.11 Crew and Payload

Flight crew : 2(pilot and co-pilot)Cabin crew : 5Passenger seating : 138 economy and 12 business class

8.12 Performance

The detailed performance estimation is given in section 9. The highlights areas follows.

• The performance is worked for a gross weight of 59175 kgf and wingloading of 5195 Nm−2 except for landing where the landing weight istaken as 85% of take-off weight.

• Maximum Mach No. at 36000 ft with cruise thrust = 0.859, with climbthrust = 0.874.

70

• Maximum still air range at M = 0.81 and h = 36000 ft is 5602 km.

• Maximum rate of climb at sea level with climb thrust = 1087 m/min

• Service ceiling = 11.55 km, Absolute ceiling = 11.88 km

• Take-off distance over 50 ft = 860 m(2820 ft) and balanced field length= 1830 m(6000 ft)

• Landing distance from 15 m = 1140 m(3740 ft)

Remark : The designed airplane meets the requirements set out in thespecifications. The seating arrangement takes care of the passenger comfortand the choice of engine reflects low level of noise.

71

9 Performance Estimation

The details regarding overall dimensions, engine details, weights, geometricparameters of wing, fuselage, nacelle, horizontal tail, vertical tail, vertical tailand other details like CLmax in various conditions and maximum load factorare given in section 8.2 - 8.10. The details of flight condition for estimationof drag polar are as follows

Altitude : 10972 m = 36000 ftMach number : 0.8Kinematic Viscosity : 3.90536 ×10−5m2/sDensity : 0.3639 kg/m3

Speed of Sound : 295.07 m/sFlight Speed : 236.056 m/sWeight of the Airplane : 59175 kgf

9.1 Estimation of Drag Polar

The drag polar is assumed to be of the form

CD = CDo +C2L

πAe

The quantity CDOis assumed to be given by

CDO= (CDO

)WB + (CDO)V + (CDO

)H + (CDO)Misc (78)

where suffices WB,V,H,Misc denote wing-body combination, vertical tail,horizontal tail, and miscellaneous contributions respectively.

9.1.1 Estimation of (CDo)WB

Initially, the drag polar is obtained at a Mach number of 0.6 as suggested by[6]. (CDo)WB is then given as :

(CDo)WB = (CDo)W + (CDo)BSBSref

The suffix B denotes fuselage and SB is the maximum frontal area of fuselage.(CDO

)W is given as :

(CDo)W = Cfw

[1 + L

(t

c

)] (SwetSref

)wing

72

Here, the Reynolds number used to determine the turbulent flat plateskin friction coefficient is based on the mean aerodynamic chord ce of theexposed wing. (Swet)e is the wetted area of the exposed wing.

Now cr = 5.59m, ct = 1.34m, b/2 = 16.11m and dfus = 3.59m. Hence

cre = 5.59− 5.59− 1.34

16.11× 3.59

2= 5.116m

λe =1.34

5.116= 0.262

ce =2

3

[5.116

(1 + 0.262 + 0.2622

1 + 0.262

)]= 3.596m

(b/2)e = 16.11− 1.795 = 14.315m

M = 0.6, a = 295.07m/s ⇒ V = 177.12m/s. Also µ = 3.90536 × 10−5.Hence,

Re =177.12× 3.596

3.90536× 10−5= 16.31× 106

k = 1.015 × 10−5m corresponds to standard camouflage paint, averageapplication (from [4]). Hence

l

k=

3.596

1.015× 10−5= 3.543× 105

The Recutoff corresponding to the above l/k is 30× 106. The Cfw is thenmeasured from the graph in [6] as

Cfw = 0.00265

(t/c)avg = 14% and (t/c)max at x/c > 0.3 ⇒ L = 1.2.

Sexposedplanform = 14.314

(5.116 + 1.341

2

)× 2 = 92.41m2

Swetw = 2× 92.41(1 + 1.2× 0.14) = 215.8m2

Hence,

(CDf)w = 0.00265 (1 + 1.2× 0.14)

215.8

111.63= 0.00598

(CDo)B is given as:

(CDO)B = (CDf

)B + (CDp)B + CDb

(CDO)B = CfB

[1 +

60

(lb/d)3 + 0.0025

(lbd

)] (SwetSB

)fus

+ CDb

SbaseSref

73

lf = 33.0m and dmax = 3.59m

Reb =177.12× 33

3.905× 10−5= 149.6× 106

k = 1.015 × 10−5m corresponds to standard camouflage paint, averageapplication. Hence

l

k=

33

1.015× 10−5= 32.51× 105

The Recutoff corresponding to the above l/k is 2.6×108. The Cfw is thenmeasured from the graph in [6] as

Cfw = 0.0019

(Swet)fus = 0.75× π × 3.59× 33 = 279m2

SB =π

4× 3.592 = 10.12m2

Hence.

(CDf)B = 0.0019× 279

10.12= 0.0524

(CDp)B = 0.0019

[60

(33/3.59)3+ 0.0025× (33/3.59)

]279

10.12= 0.00524

CDbis assumed to be zero, since base area is almost zero. Hence

(CDO)B = 0.0524 + 0.00524 + 0 = 0.0576

(∆CD)canopy is taken as 0.002. Hence (CDO)B = 0.0596

Finally we have:

(CDo)WB = 0.00598 + 0.059610.12

111.63= 0.01138

9.1.2 Estimation of (CDo)V and (CDo)H

The estimation of (CDo)H and (CDo)V can be done in a manner similar to thatfor the wing. However the details regarding the exposed tail area etc. wouldbe needed. In the absence of the detailed data on the shape of fuselage atrear etc., a simplified approach given in [6] is adopted, wherein CDf

= 0.0025for both horizontal and vertical tails.

74

SW = 2(Sh + Sv)

Hence,

(CDo)hv = 0.0025(28.71 + 25.43)1

111.63= 0.0024 (79)

9.1.3 Estimation of Misc Drag - Nacelle

For calculating drag due to the nacelles we use the short cut method forwhich we have:

(CDo)nacelle = 0.006× SwetSref

where, Swet is the wetted area of nacelle. Here Swet = 16.79m2. Since wehave two nacelles the total drag will be twice of this. Finally we get:

(CDo)nacelle = 0.006× 16.79

111.63× 2 = 0.0018

9.1.4 CDo of the airplane

Taking 2% for the interference drag (from [6]), we get the CDo of the airplaneas

CDo = 1.02 [0.01138 + 0.0024 + 0.0018] = 0.0159 (80)

9.1.5 Induced Drag

The induced drag component has the Oswald’s efficiency factor e which isestimated by adding the effect of all the aircraft components on induced drag.The rough estimate of e can be obtained from:

1

e=

1

ewing+

1

efuselage+

1

eother

From [9]ewing = (ew)Λ=0 cos(Λ− 5)

where Λ is the wing sweep. (ewing)Λ=0 = 0.97 for AR = 9.3, λ = 0.24from [12].

Hence ewing = 0.97× cos (27.69− 5) = 0.8948. Also1/efus

(Sf/S)= 0.8 for a round

fuselage. Hence

1

efus= 0.8× 10.122

111.63= 0.0725

75

1

eother= 0.05

Finally we have:

e =1

0.8948−1 + 0.0725 + 0.05= 0.8064

Hence

K =1

πAe=

1

π × 9.3× 0.8064= 0.04244

9.1.6 Final Drag Polar

CD = 0.0159 + 0.04244× C2L (81)

Figure 9: Subsonic Drag Polar

76

Remark

• The polar given by 81 is valid at subcritical Mach numbers. The in-crease in CDo and K at higher Mach numbers is discussed in section4.2.

• The maximum lift to drag ratio ((L/D)max) is given by

(L/D)max =1

2√CDoK

Using equation 81, (L/D)max is 19.25, which is typical of modern jettransport airplanes.

• It may be noted that the parabolic polar is an approximation and is notvalid beyond CLmax . It is not accurate close to CL = 0 and CL = CLmax

9.2 Engine Characteristics

To calculate the performance, the variations of thrust and SFC with speedand altitude are needed. Chapter 9 of [8] contains these variations for turbo-fan engines with various bypass ratios. The Thrust variations versus Machnumber with altitude as parameters are given in non-dimensional form fortake-off, cruise and climb ratings. The values were read from the curves andlater smoothed. The values multiplied by 97.9 kN, the sea level static thrustrating for the chosen engine, are shown in Figures 10 and 11. Figure 10also contains (a) the variation of thrust with Mach number at sea level withtake-off rating. (b) variations of climb thrust with Mach number ath = 38000 and 39000 ft; these are obtained by interpolating values at 36000and 40000 ft and are used for computation of performance at these altitudes.

The SFC variation is also given in [8], but is taken as 0.6hr−1 under cruiseconditions based on the value recommended by [4].

77

Fig

ure

10:

Tak

eO

fffo

rse

ale

velan

dC

lim

bT

hru

stper

engi

ne

for

vari

ous

alti

tudes

78

Fig

ure

11:

Cru

ise

Thru

stper

engi

ne

for

vari

ous

altitu

des

79

9.3 Level Flight Performance

In steady Level flight, the equations of motion, in standard notation are

T −D = 0 (82)

L−W = 0 (83)

L =1

2ρV 2SCL =⇒ W =

1

2ρV 2SCL (84)

D =1

2ρV 2SCD = T (85)

9.3.1 Stalling speed

In level flight,

V =

√(2W

ρSCL

)(86)

Since CL cannot exceed CLmax , there is a flight speed below which levelflight is not possible. The flight speed at CL = CLmax is called the stallingspeed and is denoted by Vs

Vs =

√(2W

ρSCLmax

)(87)

Since ρ decreases with altitude, Vs increases with height. We note thatW/S = 5195N/m2, CLmax = 2.7 with landing flaps and CLmax = 1.4 withoutflaps. The values of stalling speed at different altitudes and flap settings aretabulated in Table 6 and shown in Figure 12.

80

h ρ Vs(CLmax = 1.4) Vs(CLmax = 2.7)(m) (kg/m3) (m/s) (m/s)0 1.225 77.83 56.04

2000 1.006 85.86 61.834000 0.819 95.18 68.546000 0.659 106.06 76.378000 0.525 118.87 85.5910000 0.412 134.09 96.5611000 0.363 142.80 102.8312000 0.310 154.52 111.27

Table 6: Variation of stalling speed with altitude

Figure 12: Stalling speed Vs Altitude

81

9.3.2 Variation of Vmin and Vmax with Altitude

To determine the Vmin and Vmax at each altitude, the following procedure isadopted.

• The engine thrust as a function of velocity at each altitude is obtainedfrom the smoothed data.

• The drag at each altitude is found as a function of velocity using thedrag polar and the level flight formulae given below.

CL =2 ∗ (W/S)

ρV 2(88)

CD = CDo +KC2L (89)

Drag =1

2ρV 2SCD (90)

Tavail = f(M) (91)

Where CDo = 0.0159 and K = 0.04244.

However, the cruise Mach number (Mcruise) for this airplane is 0.8.Hence CDo and K are expected to become functions of Mach numberabove Mcruise. To get some guidelines about variations of CDo and K,we consider the drag polars of B-727 given in Volume 6, Chapter 5 of[13]. These drag polars are shown in the Figure 13 as discrete points.

82

Figure 13: Drag polars at different Mach numbers for B727-100; Symbols aredata from [13] and Solid lines are the parabolic fits

These polars were approximated by the parabolic polar expressionnamely CD = CDo +K ×C2

L. The values of CDo and K for the variousMach numbers are given in the Table 7. The parabolic fit is also shownin Figure 13.

M CDo K0.7 0.01631 0.049690.76 0.01634 0.052570.82 0.01668 0.061010.84 0.01695 0.068070.86 0.01733 0.081830.88 0.01792 0.103

Table 7: Variation of CDo and K with Mach number (Parabolic fit)

The variations in CDo and K with Mach number are plotted in theFigures 14 and 15. It is seen that there is no significant increase in

83

Figure 14: Variation of CDo with Mach number

CDo and K upto M = 0.76. This is expected to be the cruise Machnumber for the airplane (B727-100). Following analytical expressionshave been found to closely represent the changes in CDo and K fromM = 0.76 to M = 0.86.

CDo = 0.01634− 0.001× (M − 0.76) + 0.11× (M − 0.76)2 (92)

K = 0.05257 + (M − 0.76)2 + 20.0× (M − 0.76)3 (93)

In the case of the present airplane, the cruise Mach number is 0.8. Thevariations of CDo and K above Mcruise and upto M = 0.9, based onB727-100 data is taken as follows.

CDo = 0.0159− 0.001× (M − 0.8) + 0.11× (M − 0.8)2 (94)

K = 0.0455 + (M − 0.8)2 + 20.0× (M − 0.8)3 (95)

84

Figure 15: Variation of K with Mach number

• The thrust available and thrust required curves are plotted at eachaltitude as a function of velocity. The points of intersection give theVmin and Vmax at each altitude. To arrive at Vmin, the stalling speedalso needs to be taken in to account. Hence in the Figures. 16 to 21,the portion of the Vmin curve below Vs is shown as dotted lines, as thedrag polar is not valid there. Vs is taken for CLmax without flaps.

The calculations are carried out for h = 0, 10000, 15000, 25000, 30000and 36000 ft, i.e S.L, 3048, 4572, 7620, 9144 and 10972.8 m using Tavailas climb thrust and cruise thrust. Results are presented only for climbthrust case.

85

h h Vs Vmin(m/s) Vmin(m/s) Vmax(m/s) Vmax(m/s)(in ft) (in m) Tcr Tclimb Tcr TclimbS.L 0 77.833 < Vs < Vs 258.711 269.370

10000 3048 90.579 < Vs < Vs 272.060 280.59515000 4572 98.131 < Vs < Vs 275.613 283.30025000 7620 116.292 < Vs < Vs 272.929 279.29130000 9144 127.278 < Vs < Vs 267.854 271.75536000 10972 142.594 176.054 169.071 253.671 258.15438000 11582 149.557 217.386 200.896 243.676 248.63038995 11884 153.159 235.471 229.865 235.483 238.649

Table 8: Variation of Vmin and Vmax

Figure 16: Available and Required Thrust at S.L

86

Figure 17: Available and Required Thrust at h = 3048.0m

Figure 18: Available and Required Thrust at h = 4572.0m

87

Figure 19: Available and Required Thrust at h = 7620.0m

Figure 20: Available and Required Thrust at h = 9144.0m

88

Figure 21: Available and Required Thrust at h = 10972.8m

Figure 22: Variation of Vmin and Vmax with altitude

89

9.4 Steady Climb

In this flight, the C.G of the airplane moves along a straight line inclined tothe horizontal at an angle γ. The velocity of flight is assumed to be constantduring the climb. Since the flight is steady, acceleration is zero and theequations of motion can be written as:

T −D −W sin γ = 0 (96)

L−W cos γ = 0 (97)

To calculate the variation of rate of climb with flight velocity at differentaltitudes, we adopt the following procedure.

• Choose an altitude.

• Choose a flight speed.

Noting that CL = 2W cos γ/ρSV 2, we get

CD = CDo +K

(2W cos γ

ρSV 2

)Also

Vc = V sin γ

cos γ =

√1− V 2

c

V 2

Using the above equations,

A

(VcV

)2

+B

(VcV

)+ C = 0 (98)

A =kW 2

12ρV 2S

; B = −W ; C = Tavail −1

2ρV 2SCDo −

2kW 2

ρV 2S(99)

Since altitude and flight velocity have been chosen, the thrust availableis read from the climb thrust curves in 10. Further the variation of CDo

and K with Mach number is taken as in Equations 94 and 95.

90

• Equation 98 gives 2 values of Vc/V . We choose the value which is lessthat 1.0 as sin γ cannot be greater than unity. Hence

γ = sin−1(Vc/V ) (100)

Vc = V sin γ (101)

• This procedure is repeated for various speeds between Vmin and Vmax.

The entire procedure is then repeated for various altitudes.

The variations of (R/C) and γ with velocity and with altitude as pa-rameters are shown in Figure 23 and 25. The variations of (R/C)maxand γmax with altitude are shown in Figure 24 and 26. The variationsof V(R/C)max and Vγmax with altitude are shown in Figure 27 and 28. Asummary of results is presented in table 9.

h h (R/C)max V(R/C)max γmax Vγmax

(in ft) (in m) (in m/min) (in m/s) (in degrees) (in m/s)0 0.0 1086.63 149.7 8.7 88.5

10000 3048.0 867.34 167.5 6.0 111.615000 4572.0 738.16 174.0 4.7 125.725000 7620.0 487.41 198.2 2.6 164.130000 9144.0 313.43 212.2 1.5 188.036000 10972.8 115.57 236.1 0.5 230.238000 11582.4 41.58 236.9 0.2 234.038995 11885.7 1.88 235.8 0.0 235.8

Table 9: Climb Performance

91

Figure 23: Rate of Climb Vs Velocity for various altitudes

Figure 24: Maximum Rate of Climb Vs Altitude

92

Figure 25: Angle of Climb Vs Velocity for various altitudes

Figure 26: Maximum angle of Climb Vs Altitude

93

Figure 27: Velocity at Maximum Rate of Climb Vs Altitude

Figure 28: Velocity at Maximum angle of Climb Vs Altitude

94

Remarks

1. The discontinuties in slope in Figures 27 and 28 at high velocities aredue to the change in drag polar as the Mach number exceeds 0.8.

2. From Figure 24, the absolute cieling (at which (R/C)max is zero) is11.88 km. The service cieling at which (R/C)max = 50m/min is 11.55km

95

9.5 Range and Endurance

In this section, the range of the aircraft in a constant altitude and constantvelocity cruise is studied. Range is given by the formula

R =3.6V

TSFC√KCdo

[tan−1 2W1

ρV 2S

√K

Cdo

− tan−1 2W2

ρV 2S

√K

Cdo

](102)

where W1 is the weight of the aircraft at the start of the cruise and W2

is the weight of the aircraft at the end of the cruise.The cruising altitude taken is h = 10972m. TSFC is taken to be con-

stant as 0.6hr−1. The variation of drag polar above M = 0.8 is given byEquation.94 and 95.

W1 = Wo = 59175× 9.81N

Wf = 0.205×W1

Allowing 6% fuel as trapped fuel, W2 becomes

W2 = W1 − 0.94×Wf

The values of endurance (in hours) are obtained by dividing the expres-sion for range by 3.6V where V is in m/s. The values of Range(R) andEndurance(E) in flight at different velocities are presented in Table 10 andare plotted in Figures 29 and 30.

96

Figure 29: Constant Velocity Range at h = 10972 m

Figure 30: Endurance at h = 10972 m

Remarks

1. It is observed that the maximum range of 5600 km is obtained at avelocity of 239m/s (860 kmph). Corresponding Mach number is 0.81

97

M V CL CD L/D R E(in m/s) (in km) (in hours)

0.50 147.531 1.312 0.089 14.75 2979.0 5.610.55 162.285 1.085 0.066 16.48 3608.0 6.180.60 177.038 0.911 0.051 17.82 4189.6 6.570.65 191.791 0.777 0.041 18.72 4691.7 6.800.70 206.544 0.670 0.035 19.17 5095.6 6.850.75 221.297 0.583 0.030 19.23 5396.5 6.770.80 236.050 0.513 0.027 18.95 5599.8 6.590.81 239.001 0.500 0.027 18.78 5602.3 6.510.82 241.952 0.488 0.027 18.36 5527.0 6.350.83 244.902 0.476 0.027 17.65 5352.2 6.070.84 247.853 0.465 0.028 16.62 5070.1 5.680.85 250.803 0.454 0.030 15.29 4691.2 5.200.86 253.754 0.444 0.032 13.76 4242.3 4.640.87 256.705 0.433 0.036 12.13 3758.8 4.070.88 259.655 0.424 0.040 10.52 3275.3 3.50

Table 10: Range and Endurance in Constant Velocity flight at h = 10972m(36000ft)

which is slightly higher than the Mach number beyond which CDo andK increase. This can be explained based on two factors namely (i)the range increases as the flight speed increases (ii) after Mcruise isexceeded, CDo and K increase thus reducing (L/D)max.

2. The range calculated above is the gross still air range. The safe rangewould be about two-thirds of this. In the present case, the safe rangewould be 3733km.

3. The maximum endurance of 6.85 hours occurs in a flight at V =206m/s. (742 kmph). It can noted that the endurance is roughlyconstant over a speed range of 190 m/s to 230 m/s.

98

9.6 Turning Performance

In this section, the performance of the airplane in a steady, co-ordinated,level turn is studied. The equations of motion in this case are:

T −D = 0

W − L cosφ = 0

L sinφ =W

g

where φ is the angle of bank.These equations give:

r =V 2

g tanφ

ψ̇ =V

r=g tanφ

V

Load Factor n =L

W=

1

cosφ

where n = L/W , ψ̇ is the rate of turn and r is the radius of turn.The following procedure is used to obtain rmin and ψ̇max

1. A flight speed and altitude are chosen and the level flight lift coefficientCLL is obtained as :

CLL =2(W/S)

ρV 2

2. If CLmax/CLL < nmax, where nmax is the maximum load factor forwhich the aircraft is designed, then the turn is limited by CLmax andCLT1 = CLmax . However if CLmax/CLL > nmax, then the turn is limitedby nmax, and CLT1 = nmaxCLL.

3. From the drag polar, CDT1 is obtained corresponding to CLT1 . Then

DT1 =1

2ρV 2SCDT1

If DT1 > Ta, where Ta is the available thrust at that speed and alti-tude, then the turn is limited by the engine output. In this case, themaximum permissible value of CD in turning flight is found from

CDT =Ta

12ρV 2S

99

From the above relation, the value of CLT is calculated as

CLT =

√CDT − CDo

K

However if DT1 < Ta, then the turn is not limited by the engine outputand the value of CLT calculated in step (ii) is retained.

4. Once CLT is known, the load factor during the turn is determined as

n =CLTCLL

Once n is known, the values of φ, r and ψ̇ can be calulated using theequations given above.

The above steps are then repeated for various speeds and altitudes. Atypical turning flight performance estimation is presented in Table 11. Inthese calculations, CLmax = 1.4 and nmax = 3.5 are assumed. The variationof turning flight performance with altitude is shown in Table. 12. Figures31, 32, 33, 34 respectively present (a) radius of turn with velocity and withaltitude as parameter, (b) minimum radius of turn with altitude, (c) rate ofturn with velocity and with altitude as parameter and (d) maximum rate ofturn with altitude.

v n Clt φ (in degrees) r (in m) ψ̇78.83 1.026 1.4000 12.892 2767.70 0.028598.83 1.612 1.4000 51.670 787.21 0.1255118.83 2.331 1.4000 64.596 683.63 0.1738138.83 2.813 1.2376 69.173 747.41 0.1858158.83 2.993 1.0062 70.482 911.60 0.1742178.83 3.089 0.8192 71.112 1115.38 0.1603198.83 3.080 0.6607 71.053 1383.50 0.1437218.83 2.930 0.5189 70.045 1772.43 0.1235238.83 2.573 0.3826 67.132 2452.36 0.0974241.83 2.494 0.3617 66.363 2609.20 0.0927

Table 11: A typical turning flight performance at Sea level

100

Figure 31: Radius of Turn Vs Velocity at various altitudes

Figure 32: Velocity at Rmin Vs Altitude

101

Figure 33: ψ̇ Vs Speed at various altitudes

Figure 34: Velocity at ψ̇max Vs Altitude

102

h rmin Vrminψ̇max Vψ̇max

(in m) (in m) (in m/s) (in m/s)0.0 666 126.8 0.1910 127.8

3048.0 945 132.6 0.1410 133.64572.0 1155 135.1 0.1170 136.17620.0 1971 138.3 0.0731 165.39144.0 3247 151.3 0.0513 187.310972.8 8582 211.0 0.0256 231.0

Table 12: Turning flight performance

Remarks

1. The maximum value of ψ̇ is 0.191 and occurs at a speed of 127.8m/sat sea level.

2. The minimum radius of turn is 666 m and occurs at a speed of 126.8m/sat sea level.

3. The various graphs show a discontinuity in slope when the criterionwhich limits the turn changes from nmax to thrust available.

9.7 Take-off distance

In this section, the take off performance of the airplane is evaluated. Thetake-off distance consists of take-off run, transition and climb to screenheight. Rough estimates of the distance covered in these phases can beobtained by writing down the appropriate equations of motion. However theestimates are approximate and [4] recommends the following formulae fortake-off distance and balance field length based on the take-off parameter.This parameter is defined as:

Take Off Parameter =W/S

σCLTO(T/W )

(103)

where W/S is wing loading in lb/ft2, CLTOis 0.8× CLland = 0.8× 2.7 =

2.16 and σ is the density ratio at take-off altitude.In the present case:

W

S= 5195N/m2 = 108.2lb/ft2; CLTO

= 0.8×2.7 = 2.16; σ = 1.0(sea level)

103

andT

W=

2× 97.9kN

59175× 9.81= 0.3373

Hence

Take Off Parameter =108.2

1.0× 2.16× 0.3373= 148.86 (104)

From [4], the take off distance, over 50’, is 2823′ or 861m. The balancefield length for the present case of two engined airplane is 6000′ or 1829m.

Remark

It may be noted that the balance field length is more than twice the take offdistance itself.

9.8 Landing distance

In this section the landing distance of the airplane is calculated. From [4]the landing distance for commericial airliners is given by the formula

Sland = 80

(W

S

)1

σCLmax

+ 1000ft (105)

where W/S is in lbs/ft2. In the present case:

• (W/S)land = 0.85× (W/S)takeoff = 0.85× 108.5 = 92.225lb/ft2

• CLmax = 2.7

• σ = 1.0

Hence

Sland = 80× 92.2251

1.0 ∗ 2.7+ 1000 = 3732ft = 1138m (106)

104

9.9 Concluding remarks

1. Performance of a typical commercial airliner has been estimated forstalling speed, maximum speed, minimum speed, steady climb, range,endurance, turning, take-off and landing.

2. The performance approximately corresponds to that of B737-200.

3. Figure 35 presents the variation with altitude of the characteristic ve-locities corresponding to

• stalling speed, Vs

• maximum speed, Vmax

• minimum speed as dictated by thrust, Vminthrust

• maximum rate of climb, V(R/C)max

• maximum angle of climb, Vγmax

• maximum rate of turn, Vψ̇max

• minimum radius of turn, Vrmin

105

Fig

ure

35:

Fligh

tE

nve

lope

106

10 Acknowledgements

The first author(EGT) thanks AICTE for the fellowship which enabled himto carry out the work at IIT Madras.

References

[1] http://www.cfm56.com/engines/cfm56-5c/tech.html

[2] http://www.lissys.demon.co.uk/samp1/

[3] NASA Technical Paper 2969, Charles Harris (Mar 1990)

[4] Raymer.D.P. Aircraft design a conceptual approach. AIAA’ educationalseries, 2006

[5] Tulapurkara.E.G Lecture Notes on Aircraft Design, Department ofAerospace Engineering I.I.T Madras, 2007

[6] Roskam J. Methods of estimating drag polars of subsonic air-planesRoskam Aviation & Engineering Corporation, Ottawa, Kansas,1983

[7] Lebedenski.A.A Aircraft design parametric studies Published by I.I.Sc,Bangalore, 1971

[8] Jenkinson L.R., Simpkin P. and Rhodes D. Civil Jet Aircraft Design,Arnold, 1999

[9] Hoerner S.F. Fluid dynamic drag, published by Hoerner Fluid Dynamics,Brick Town, NJ, 1965

[10] Wood K.D. Aerospace vehicle design, Volume 1, Johnson publishingcompany, Boulder, Colorado, 1966

[11] Perkins C.D. & Hage A.E. Airplane performance syability & control,McGraw Hill, 1963

[12] Abbot I.H. and Doenhoff A.E. Theory of wing sections, Dover publica-tions, 1959

[13] Roskam J. Aircraft design, Roskam Aviation & Engineering Corpora-tion, Ottawa, Kansas, 1990

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