Aircraft Design Project - 150 seater passenger aircraft

April 22, 2017 | Author: sonusingh7867 | Category: N/A
Share Embed Donate


Short Description

Its a aircraft design project (ADP) of a 150 seater passenger aircraft....

Description

AIRCRAFT DESIGN PROJECT-1 150 SEATER PASSENGER AIRCRAFT

SUBMITTED BY: VELURU VENKATA RAMANA VEDICHERLA VAMSI KRISHNA VISWANADULA ADI SESHU

ACKNOWLEDGEMENT

I would like to extend my heartful thanks to Prof. ASHOKAN (Head of Aeronautical Department) for giving me his able support and encouragement. At this juncture I must emphasis the point that this DESIGN PROJECT would not have been possible without the highly informative and valuable guidance by Prof.SARVESWARAN, whose vast knowledge and experience has must us go about this project with great ease. We have great pleasure in expressing our sincere & whole hearted gratitude to them. It is worth mentioning about my team mates, friends and colleagues of the Aeronautical department, for extending their kind help whenever the necessity arose. I thank one and all who have directly or indirectly helped me in making this design project a great success.

3|Page

INDEX Serial No. Topic

Page No.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

5 7 9 16 20 39 41 49 53 55 60 70 75 81 87 94

17 18 19

Aim of the Project Abstract Introduction Comparative DataSheet Graphs Mean Design Parameters Weight Estimation Powerplant Selection Fuel Weight Validation Wing Selection Airfoil Selection Lift Estimation Drag Estimation Landing Gear Arrangement Fuselage Design Performance Characteristics 3 – View Diagram Conclusion Bibliography

100 104 106

4|Page

ABBREVIATION A.R. - Aspect Ratio B - Wing Span (m) C - Chord of the Airfoil (m) C root - Chord at Root (m) C tip - Chord at Tip (m) - Mean Aerodynamic Chord (m) C Cd - Drag Co-efficient Cd,0 - Zero Lift Drag Co-efficient Cp - Specific fuel consumption (lbs/hp/hr) CL - Lift Co-efficient D - Drag (N) E - Endurance (hr) E - Oswald efficiency L - Lift (N) (L/D)loiter - Lift-to-drag ratio at loiter (L/D)cruise - Lift-to-drag ratio at cruise M - Mach number of aircraft Mff - Mission fuel fraction R - Range (km) Re - Reynolds Number S - Wing Area (m²) Sref - Reference surface area Swet - Wetted surface area Sa - Approach distance (m) Sf - Flare Distance (m)

5|Page

Sfr - Free roll Distance (m) Sg - Ground roll Distance (m) T - Thrust (N) Tcruise - Thrust at cruise (N) Ttake-off - Thrust at take-off (N) (T/W)loiter - Thrust-to-weight ratio at loiter (T/W)cruise - Thrust-to-weight ratio at cruise (T/W)take-off - Thrust-to-weight ratio at take-off Vcruise - Velocity at cruise (m/s) Vstall - Velocity at stall (m/s) Vt - Velocity at touch down (m/s) Wcrew - Crew weight (kg) Wempty - Empty weight of aircraft (kg) Wfuel - Weight of fuel (kg) Wpayload - Payload of aircraft (kg) W0 - Overall weight of aircraft (kg) W/S - Wing loading (kg/m²) ρ - Density of air (kg/m³) μ- Dynamic viscosity (Ns/m²) λ - Tapered ratio R/C - Rate of Climb

6|Page

AIM OF THE PROJECT The aim of this design project is to design a150 seater passenger aircraft by comparing the data and specifications of present aircrafts in this category and to calculate the performance characteristics. Also necessary graphs need to be plotted and diagrams have to be included wherever needed. The following design requirements and research studies are set for the project:  Design an aircraft that will transport 150 passengers and their baggage over a design range of 4820 km at a cruise speed of about 890 km/h.  To provide the passengers with high levels of safety and comfort.  To operate from regional and international airports.  To use advanced and state of the art technologies in order to reduce the operating costs.

7|Page

 To offer a unique and competitive service to existing scheduled operations.  To assess the development potential in the primary role of the aircraft.  To produce a commercial analysis of the aircraft project.

8|Page

ABSTRACT The purpose of the project is to design a 150 seater Medium Range International passenger aircraft. The aircraft will possess a low wing, tricycle landing gear and a conventional tail arrangement. Such an aircraft must possess a wide body configuration to provide sufficient seating capacity. It must possess turbofan engines to provide the required amount of speed, range and fuel economy for the operator. The aircraft will possess two engines.

9|Page

INTRODUCTION At the instant time there are different types of aircrafts with latest technology. Every year there is a great competition for making an aircraft of having higher capacity of members inside the aircraft. So here in this report, we intend to implant the differentiation among the aircrafts having sitting capacity of 100-180 members. This report gives the different aspects of specifications like wing specification, weight specification, power plant specification and performance specification. Airbus started the development of a very large airliner (termed Megaliner by Airbus in the early development stages) in the early 1990s, both to complete its own range of products and to break the dominance that Boeing had enjoyed in this market segment since the early 1970s with its 747. McDonnell Douglas pursued a similar strategy with its ultimately unsuccessful MD-12 design. As each manufacturer looked to build a successor to the 747, they knew there was room for

10 | P a g e

only one new aircraft to be profitable in the 600 to 800 seat market segment. Each knew the risk of splitting such a niche market, as had been demonstrated by the simultaneous debut of the Lockheed L-1011 and the McDonnell Douglas DC-10: both planes met the market’s needs, but the market could profitably sustain only one model, eventually resulting in Lockheed's departure from the civil airliner business. In January 1993, Boeing and several companies in the Airbus consortium started a joint feasibility study of an aircraft known as the Very Large Commercial Transport (VLCT), aiming to form a partnership to share the limited market. Airplanes come in many different shapes and sizes depending on the mission of the aircraft, but all modern airplanes have certain components in common. These are the fuselage, wing, tail assembly and control surfaces, landing gear, and powerplant. For any airplane to fly, it must be able to lift the weight of the airplane, its fuel, the passengers, and the cargo. The wings generate most of the lift to hold the plane in the air. To generate lift, the airplane must be pushed through the air. The engines, which are usually located beneath the wings, provide the thrust to push the airplane forward through the air.

11 | P a g e

The fuselage is the body of the airplane that holds all the pieces of the aircraft together and many of the other large components are attached to it. The fuselage is generally streamlined as much as possible to reduce drag. Designs for fuselages vary widely. The fuselage houses the cockpit where the pilot and flight crew sit and it provides areas for passengers and cargo. It may also carry armaments of various sorts. Some aircraft carry fuel in the fuselage; others carry the fuel in the wings. In addition, an engine may be housed in the fuselage.

The wing provides the principal lifting force of an airplane. Lift is obtained from the dynamic action of the wing with respect to the air. The cross-sectional shape of the wing as viewed from the side is known as the airfoil section. The planform shape of the wing (the shape of the wing as viewed from above) and placement of the wing on the fuselage (including the angle of incidence), as well as the airfoil section shape, depend upon the airplane mission and the best compromise necessary in the overall airplane design.

12 | P a g e

The control surfaces include all those moving surfaces of an airplane used for attitude, lift, and drag control. They include the tail assembly, the structures at the rear of the airplane that serve to control and maneuver the aircraft and structures forming part of the tail and attached to the wing.

13 | P a g e

PURPOSE AND SCOPE OF AIRPLANE DESIGN OBJECTIVES  To meet the FUNCTIONAL, OPERATIONAL and SAFETY requirements set out OR acceptable to the USER. ACTUAL PROCESS OF DESIGN    

Selection of aircraft type and shape Determination of geometric parameters Selection of power plant Structural design and analysis of various components  Determination of aircraft flight and operational characteristics .

14 | P a g e

DESIGN CYCLE PRELIMINARY DESIGN It consists of the initial stages of design, resulting in the presentation of a BROCHURE containing preliminary drawings and clearly stating the operational capabilities of the airplane being designed. This Brochure has to be APPROVED by the manufacturer and/or the customer. The steps involved:  Layout of the main components  Arrangement of airplane equipment and control systems  Selection of power plant  Aerodynamic and stability calculations  Preliminary structural design of MAJOR components  Weight estimation and c.g. travel  Preliminary and Structural Testing  Drafting the preliminary 3-view Drawings

15 | P a g e

DESIGN PROJECT             

Internal discussions Discussions with prospective customers Discussions with Certification Authorities Consultations with suppliers of power plant and major accessories Deciding upon a BROAD OUTLINE to start the ACTUAL DESIGN, which will consist of Construction of Mock-up Structural layout of all the individual units, and their stress analysis Drafting of detailed design drawings Structural and functional testing Nomenclature of parts Supplying key and assembly diagrams Final power plant calculations Final weight estimation and c.g. limits Final performance calculation

16 | P a g e

CYCLES OF DESIGN PROCESS:

17 | P a g e

Aircraft design can be broken into three major phases,

18 | P a g e

(a) Conceptual Design (b) Preliminary Design (c) Detail Design

CONCEPTUAL DESIGN: Conceptual design is a very fluid process. New ideas and problems emerge as a design is investigated in ever increasing detail. Each time the latest design is analyzed and sized, it must be redrawn to reflect the new gross weight, fuel weight, wing size, engine size, and other changes. Conceptual design will usually begin with either a specific set of design requirements established by the prospective customer or a company –generated guess as to what future customers need. Design requirements include aircraft range and payload, takeoff and landing distances, and maneuverability and speed requirements.

19 | P a g e

The actual design effort usually begins with conceptual sketch. A good conceptual sketch will include the approximate wing and tail geometries, the fuselage shape, and the internal locations of the major components

such

as

the

engine,

cockpit,

payload/passenger compartment, landing gear and fuel tanks.

PRELIMINARY DESIGN: It can be said to begin when the major changes are over. The big questions such as whether to use a canard or an aft tail have been resolved. At some point late in preliminary design, even minor changes are stopped when a decision is made to freeze the

20 | P a g e

configuration. During this design the specialists in areas such as structures, landing gear, and control systems will design and analyze their portion of the aircraft. Testing is initiated in areas such as aerodynamics, propulsion, structures, and stability and control. A key activity during this type of design is “LOFTING’. Lofting is the mathematical modeling of the outside skin of the aircraft with sufficient accuracy to insure proper fit between its different parts, even if they are designed by different designers and possibly fabricated in different locations. The ultimate objective during this design is to ready the company for the detail stage, also called “FULL-SCALE DEVELOPMENT”.

DETAIL DESIGN:

21 | P a g e

Assuming a favorable decision for entering full-scale development, the detail design phase begins in which the actual pieces to be fabricated are designed. For example, during conceptual and preliminary design the wing box will be designed and analyzed as a whole. During detail design, that whole will be broken down into individual ribs, spars, and skins, each of which must be separately designed and analyzed. Another important part of detail design is called production design. Specialists determine how the airplane will be fabricated, starting with smallest and simplest subassemblies and building upto the final assembly process. Production designers frequently wish to modify the design for ease of manufacture; that can have a major impact on performance or weight. Compromises are inevitable, but the design must still meet the original requirements.

22 | P a g e

During detail design, the testing effort intensifies. Actual structure of the aircraft is fabricated and tested. Control laws for the flight control system are tested on an “iron-bird” simulator, a detailed working model of the actuators and flight control surfaces. Flight simulators are developed and flown by both company and customer test pilots. Detail design ends with fabrication of the aircraft. Frequently the fabrication begins on part of the aircraft before the entire detail-design effort is completed.

Comparative Datasheet – 1

23 | P a g e

NAME OF AIRCRAFT

Boeing

Boeing C-40 Boeing

Boeing

737-100

Clipper

717-200

737-200

CAPACITY

124

LENGTH (M)

28.65

WING SPAN (M)

28.35

HEIGHT (M)

11.23

WING AREA(m^2)

102

102

THRUST (kN)

64

77

EMPTYWEIGHT(kg) 28100 MAX

TAKE

121 33.3 34.2 12.55

57150

OFF 50300

WEIGHT

136

117

30.53

37.77

28.35

28.48

11.23

8.87

82.3

31600 52400

49900

10700

2645

78000

SERVICE SEILING (m) 10670 RANGE (km)

2850

ASPECT RATIO

8

12500 5600

4300 8

8.5

ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) POWER PLANT

0.407 786

839.12

P&T JT8D Cfm56-7

Comparative Datasheet – 2

786

817.867

P&T JT8D

RR BR715A1

24 | P a g e

NAME OF AIRCRAFT

AIRBUS

BOEING

Boeing

Boeing

A 318-100 737-300

737-500

737-600

CAPACITY

132

149

132

140

LENGTH (M)

31.44

33.414

31.008

31.2

WING SPAN(M)

34.1

28.9

28.9

35.8

HEIGHT(M)

12.51

11.15

11.15

12.53

90

90

101

EMPTY WEIGHT(Kg) 39500

32700

31300

36378

MAX

OFF 68200

62800

60550

66000

SERVICE SEILING(M) 11887

11277

11277

12500

WING AREA(M^2) 112.6 THRUST( kN)

TAKE

106

WEIGHT (Kg)

RANGE

5700

4200

4444

5970

ASPECT RATIO

10

9.11

9.46

9.45

786

786

833.8

ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 828.488 POWER PLANT

CMF56-5

CMF56-7

25 | P a g e

Comparative Datasheet – 3 NAME OF AIRCRAFT

Boeing

Boeing

ANTONAV

COMAC

737-700

717-200

AN-10

ARJ 21

CAPACITY

148

117

100

105

LENGTH (M)

33.63

37.8

34

36.36

WING SPAN (M)

35.8

28.47

38

36.35

HEIGHT (M)

12.55

8.92

9.8

8.44

84.5

121

80

WING AREA (M^2) THRUST (kN)

117

82.3

EMPTY WEIGHT (Kg) 38147

30618

MAX TAKE OFF 66000

49900

82.3 65700

26300 43616

WEIGHT (Kg) SERVICE SEILING (M) 12500

11000

11000

11900

RANGE (kM)

6370

2645

2532

2200

ASPECT RATIO

8

7.8

7

7.9

817.9

734.3

827.7

ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 833.8 POWER PLANT

CMF56-7

RR BR715

GE CF34

26 | P a g e

Comparative Datasheet – 4 NAME OF AIRCRAFT FOKKER100 FOKKER100 Boeing

Boeing

TAY620

TAY650

707-020

CAPACITY

122

122

140

179

LENGTH (M)

35.53

35.53

41.25

44.07

WING SPAN (M)

28.08

28.08

39.9

39.9

HEIGHT (M)

8.5

8.5

12.65

12.93

WING AREA (M^2) 93.5

93.5

THRUST (kN)

67.2

61.6

77-120B

EMPTY WEIGHT(Kg) 24375

24541

46785

55580

MAX TAKE OFF 43090

45810

100800

116570

WEIGHT (Kg) SERVICE SEILING(M) 11000

11000

RANGE (KM)

2450

3170

7040

8704

ASPECT RATIO

8.5

8.5

11

11

CRUISESPEED(km/h) 828

828

1005

1005

RR MK650

P&W JTD1

P&W JTD1

ENDURANCE WING LOADING THRUST TO WEIGHT RATIO

POWER PLANT

RR MK620

Comparative Datasheet – 5

27 | P a g e

NAME OF AIRCRAFT Boeing

Boeing

Boeing

707-320B

727–100

727–200

CAPACITY

147

149

189

99

LENGTH (M)

46.61

40.6

46.7

28.91

WING SPAN (M)

44.42

32.9

32.9

28.91

HEIGHT (M)

12.93

10.3

10.3

8.6

WING ARE (Sq.M)

Antonov An-158

87.32

THRUST (kN)

67.0

EMPTY WEIGHT (Kg) 66406 MAX TAKE OFF 151320

45360

45360

76818

95028

WEIGHT (Kg) SERVICE SEILING RANGE (kM)

10650

5000

4400

ASPECT RATIO

11

9

9

6.9

862

862

798

ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 968.4 POWER PLANT

PW JTD-3D PW JT8D-7

D-36

28 | P a g e

COMPARITIVE GRAPHS:

29 | P a g e

LENGTH: 3o

RANGE:4800KM

30 | P a g e

ASPECT RATIO:9.3

31 | P a g e

MAXIMUM TAKE OFF WEIGHT:38506 kg

32 | P a g e

EMPTY WEIGHT:31620 kg

33 | P a g e

FUEL WEIGHT:10405 kg

34 | P a g e

SERVICE CELING:11640 m

35 | P a g e

THRUST:60 kN

36 | P a g e

WING LOADING:547 Kg/Sq.M

37 | P a g e

WING SPAN:27.2M

38 | P a g e

WING AREA:80 M^2

39 | P a g e

DESIGN DATA SHEET

GENERAL CHARACTERISTICS CREW

4

PASSENGER CAPACITY

100-180

LENGTH

30 m (98ft 5 inch)

WING SPAN (b)

27.2 m (89ft 2 inch)

WING AREA (S)

80.1 m2 (862.18 ft2)

ASPECT RATIO(b2/S)

9.3

MAX TAKE OFF WEIGHT

38,506 kg (84,891 lb)(corrected)

EMPTY WEIGHT

31620 kg (69,710 lb)

FUEL WEIGHT

10,405 kg (22,939 lb)

POWERPLANT

ROLLS ROYCE TAY 620-15 TURBOFAN

NO OF ENGINES

2

40 | P a g e

THRUST

60 KN (13,488 lbf)

PERFORMANCE CHARACTERISTICS MAX SPEED

940 km/h (0.78 mach, 584 mph)

CRUISE SPEED

890 km/h (0.74 mach,553 mph)

RANGE

4,820 km (n 2,995 miles)

SERVICE CEILING

11,640 m(38,188 ft)

MISSION SPECIFICATION

The aircraft undergoes a simple mission. The need of fuel is necessary only for the shown cruise. The reserve fuel as mentioned above is needed during loiter.

41 | P a g e

PAYLOAD: 150 passengers at 75kg each and 25kg of baggage each. CREW: 2 pilots and 2 cabin attendants at 200lbs and 30lbs baggage each respectively. RANGE: 2502nm, followed by 1 hour loiter followed by a 100nm flight to alternate. ALTITUDE: 33,100 feet (For the design range). CRUISE SPEED: M=0.74, at 38,100 ft. CLIMB: A dir6-ect climb to 33,100 ft. at Max. WTO is desired. POWERPLANTS: 2 Turbo-fan engines.

MISSION SPECIFICATION

The aircraft undergoes a simple mission. The need of fuel is necessary only for the shown cruise. The reserve fuel as mentioned above is needed during loiter.

42 | P a g e

PAYLOAD: 150 passengers at 75kg each and 25kg of baggage each. CREW: 2 pilots and 2 cabin attendants at 75kgs and 25kgs baggage each respectively. RANGE: 2502nm, followed by 1 hour loiter followed by a 100nm flight to alternate. ALTITUDE: 33,100 feet (For the design range). CRUISE SPEED: M=0.74, at 38,100 ft. CLIMB: A direct climb to 33,100 ft. at Max. WTO is desired. POWERPLANTS: 2 Turbo-fan engines.

43 | P a g e

44 | P a g e

WEIGHT ESTIMATION

Airplane must normally meet very stringent range, endurance, speed and cruise speed objectives while carrying a given payload. It is vital in predicting the minimum airplane weight and fuel weight needed to accomplish a given mission. We know the lift acquired is directly proportional to the weight of the aircraft. Hence the estimation of weight for a given aircraft plays a key role in design analysis.

W= Wstructure + Wpayload + Wfuel + Wcrew +Wpower plant + Wfixed equipment

45 | P a g e

Payload Weight Validation (WPL): WPL =

No. of passengers*(Wt.of passenger+Wt.of

baggage) = WPL =

150*(75+25) 15000 kg

Crew Weight Validation (WCREW): WCrew = WCrew

=

No. of crew members*(Crew wt. +Baggage)

4*(75+10) =

340 kg

Approximate Take-off Weight(WTO(approx)):

46 | P a g e

WTO(approx)=

38393 kg [From Design data sheet]

Now, taking into consideration the appropriate mission phases:

P a g e | 47

Airplane Take Off Climb Type Business Jets Transpor t Military Trainers Superson ic Cruise

Descent

Landing

0.995

0.980

0.990

0.992

0.970

0.985

1.000

0.995

0.990

0.980

0.990

0.995

0.995

0.92-0.87 0.985

0.992

For takeoff, segment 0-1 historical data’s shows that, 𝑊1/𝑊0 = 0.97.

P a g e | 48

For climb, segment 1-2 historical data shows that, W2 = 0.985 W1

Aircraft Design Project - 1 For loiter, segment 3-4 ignoring the fuel consumption during descent we assume, 𝑊4/𝑊3 = 1

For landing, segment 4-5 based on historical data we assume that, 𝑊5/𝑊4 = 0.995

For landing, segment 4-5 based on historical data we assume that, 𝑊5/𝑊4 = 0.995

P a g e | 49

The Brequet’s range equation is used to calculate the value of 𝑤3𝑤2. As we all know that maximum range is covered during cruise we considering this equation,

R = (𝑣∞/𝑐𝑗)(𝐿/𝐷)ln (𝑤2/𝑤3)

L/D values of similar type of aircrafts we come to know that the approximate the value of L/D for our aircraft to be 15. So, 𝐿/𝐷 =15 From the comparative data sheet, V∞ = 872 km/hr R = 7200 km

For Business & transport jets We found the values of cj as 0.6 hr-1

P a g e | 50

So now substituting these values in the Brequet’s range equation, R = (𝑣∞/(𝐿/𝐷)(ln 𝑤2/𝑤3) = (940/.6)(15)ln(W2/W3) ln(w2/w3)=(4820/23500)= 0.225 (w2/w3)=1.25 W3/w2=0.8

Mission Fuel Fraction Weight Validation (MFF):

The fuel fraction of each phase is defined as the ratio of the end weight to the begin weight. MFF = (w1/w0)x(w2/w1)x(w3/w2)x(w4/w3)x(w5/w4) = 0.97x0.985x0.8x1x0.995

P a g e | 51

MFF =

0.796

Fuel Weight Validation (WFUEL):

WFUEL = WFUEL

=

(1-Mff)* WTO appro

(1-0.796)*38393 =

7832.2 kg

Approximate Operational Weight Validation (WOE(approx)): WOE(apporx ) = = WOE(apporx ) =

WOE(apporx)- WFUEL- WPL 38393-7832-9180 21381 kg

Tentative Empty Weight Validation (WE(tent)):

P a g e | 52

WE(tent)

=

WOE(approx) -WTFO - WCREW

Where , WTFO = = WE(tent)

0.5% of WTO(approx) 21381-0.005*38393-378 =

20811 kg

Maximum Takeoff Weight Validation (WTO):

WTO = = WTO =

WE(tent)+ WFUEL+ WPL+ WCREW 20811+7832.2+9180+378 38201.134 [CORRECTED]

P a g e | 53

Aircraft Empty Weight Validation (WE) can be further split up into:

WE =

Wstruc +WPP + WFE

We estimate the Gross Wt. (Wg) =

0.9* WTO =34381 kg

From the Group Weight Jet Transport data, Wstruc =0.321 Wg Hence

Wstruc = 11061.45 kg

WFE

= 0.169

Wg

P a g e | 54

Hence

WFE = 5828 kg

WPP = 0.114 Wg

Hence WPP = 3921.5 kg

Hence the empty weight

WE =

WE =

=

Wstruc +WPP + WFE

11061.45+3921.54+5828.15 20811 kg

P a g e | 55

POWERPLANT SELECTION

As estimated from the design data sheet, the aircraft to be designed requires a thrust of 60kN.A consolidated list of the various engines that tally with the thrust required for the given aircraft is tabulated below. Engine

Dry

Name

weight

SFC (Kg/kN.hr)

Max Thrust (kN)

P a g e | 56

(kg) RollsRoyce

1501

69.93

61.60

1200

71.2

64.54

1595

69.93

67.16

2104

72.97

66.72

1856

77.02

54.00

TAY 620 GE-CF348 RollsRoyce TAY 650 RollsRoyce BR710-48 RollsRoyce

P a g e | 57

Hence, the optimum choice of engine, from those listed above would be the Rolls-Royce TAY 620 engine which meets the demand of weight and thrust required at similar payload such as the one under design. In our design, we select 2 rear engines, namely Rolls-Royce TAY 620 Turbo Fan Engine to meet the given design standards.

Rolls-Royce TAY 620 Description

P a g e | 58

Manufacturer

ROLLS-ROYCE

Type of the engine

TAY -620 TURBOFAN

Thrust at SL/ISA

61.6 kN (13850 lbs)

Inlet mass flow

184 Kg/s(408 lb/s)

Bypass Ratio

3.04

Overall Pressure Ratio

16.0

Fan diameter

1.17 m(44 inch)

Engine weight

14220.57 Kg(3135lbs)

Engine length

2.4 m(94.8 inch)

Turbine entry Temperature

1305 K

Specific fuel consumption

69.93Kg/kN.hr(0.69 lb/lbf.hr)

(SFC) Fuel consumption(Eg 300 NM 2063 KG flight)

P a g e | 59

ROLLS-ROYCE TAY620

P a g e | 60

FUEL WEIGHT VALIDATION The choice of a suitable engine, having been made, it is now possible to estimate the amount of fuel required for a flight at the given cruising speed for the given range. Wfuel = (𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒏𝒈𝒊𝒏𝒆𝒔∗𝑻𝒉𝒓𝒖𝒔𝒕 𝒂𝒕 𝒂𝒍𝒕𝒊𝒕𝒖𝒅𝒆∗𝑹𝒂𝒏𝒈𝒆∗𝑺𝑭𝑪∗𝟏.𝟐)/ 𝑪𝒓𝒖𝒊𝒔𝒆 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 The factor of 1.2 is provided for reserve fuel. Thrust at altitude is calculated using the relation: =Ρalt/ ρo Altitude = 10800m = 35433ft 𝜍 = 𝜌 𝑎𝑙𝑡 𝜌0 = 0.3715/1.225 = 0.303 Cruise velocity = 872km/hr = 242.2m/s To = 320kN 𝑇𝜍= 320×0.3031.2 𝑇𝜍= 76.363kN = 7784.2kg SFC = 0.4hr-1 (at medium thrust setting)

P a g e | 61

Number of engines = 3

CALCULATION: Wfuel = 3×7784.2×7200×0.4×1.2 872 Wfuel = 92,553.42 kg

WING SELECTION AERODYNAMICS

Wing Configuration:

In the wing design, we have two considerations: 1. The geometric shape of the wing 2. Determination of mean aerodynamic chord 3. Wing location relative to the fuselage

P a g e | 62

GEOMETRY OF THE WING:

Wing geometry is described by

a. Plan-form shape b. Aspect ratio (which is already obtained from comparative graphs) c. Wing sweep d. Taper ratio e. Aerofoil shape and thickness along the span f. Geometric twist (change in aerofoil chord incidence angle along the span).

The maximum design velocity is 940 km/h (.78 M) - below the transonic region. We choose to use a swept wing.

Initially the primary aerodynamics data was obtained from the design data sheet. The parameters known include:

P a g e | 63

Wing span, b = 27.2m Wing Area, S = 80.1 m^2 Aspect ratio, AR = b^2/S = 9.3

Taper Ratio λ = 0.36 (from Roskam book – BAE 196-200)

We know that taper ratio,

λ = Ct = 0.36Cr

Where, Ct is the root chord Cr is the tip chord Ct =0.36 Cr Also S/2 = b/2 (Ct + Cr )/2 80.1/2

= 27.2/2 ((0.36 Cr + Cr ) / 2)

P a g e | 64

Hence

Cr = 4.33m

Also Ct = 1.56m A leading edge sweep angle of 15°. Hence we get the plan form as shown

P a g e | 65

DETERMINATION OF THE MEAN AERODYNAMIC CHORD:

Mean chord

= {(2/3)*Cr (1+λ+ λ2)}/ (1+λ)

= 3.16 m

Distance of the mean chord = {b*(1+2 λ)}/{6*(1+λ)} from the aircraft centre line = 5.73 m

P a g e | 66

RELATIVE LOCATION OF THE WING:

There are three basic vertical locations of the wing relative to the fuselage: 1. High wing 2. Mid wing 3. Low wing

High wing: 1. Low-slung fuselage – ease to place the fuselage lower to the ground. 2. More stable in lateral and rolling motion. 3. It is a distinct advantage for transport plane since it simplifies the loading and unloading processes.

P a g e | 67

Mid wing: 1. Least interference drag. 2. Gives best stability with little dihedral.

Low wing: 1. Landing gear can easily be retracted into the wing box. 2. Added fillet will avoid undesirable aerodynamic interference.

In light of all the above considerations, we choose a highwing configuration mainly due to structural and Landing gear considerations.

SELECTION OF AEROFOIL:

P a g e | 68

In order to select an aerofoil the appropriate thickness to chord ratio(t/c) is to be determined.

The given formula can be used to determine the thickness to chord ratio: Volume of fuel = 2*{((2/3)*(t/c)*c)*c*(b/4)}*0.5*1.33

Volume of fuel = 2*{((2/3)*(t/c)*c)*c*(b/4)}*2

Volume of fuel =

Weight of fuel(kg) Specific gravity of fuel

=

7894.716 (0.72*1000)

= 10.965 m^3

P a g e | 69

Substituting the above weight of fuel in the equation and solving, we get the value of thickness to chord ratio as

10.965

=

2*{((2/3)*(t/c)*2.945)*2.945*(27.2/4)}*0.5*1.33 t/c = 0.209 (or) 0.21 As our aircraft will fly only at subsonic speeds, we have chosen a NACA 6 series aerofoil. Based on the t/c ratio, we have chosen the aerofoil NACA 664-221.

P a g e | 70

FUSELAGE LAYOUT The fuselage layout is important as the length of the entire aircraft depends on this.

The length and diameter of the fuselage is related to the seating arrangement.

The fuselage of a passenger aircraft is divided into a number of a sections:

1. Nose 2. Cockpit 3. Cabin 4. Tail fuselage

P a g e | 71

Functions of fuselage:  provision of volume for payload  provide overall structural integrity  possible mounting of landing gear and power plant

Once fundamental configuration is established, fuselage layout proceeds almost independent of other design aspects Pressurisation  If required, has a major impact upon the overall shape.  Overall effect depends on the level of pressurization. Fuselage sizing:

P a g e | 72

A relation between the gross weight of the aircraft and the length of the aircraft. Lfus = aWb Where W is in lbs and Lfus is in ft. For jet transport,a=0.67,b=0.43 Lfus = 19.5m

Nose and cockpit-Front fuselage The layout of the flight deck and specified pilot window geometry is often the starting point of the overall fuselage layout.

For the current design, flight decks of various airplanes are considered and the value of Lnose/Lfus is found to be 0.03

Lnose = 0.03*19.5 = 0.58m

P a g e | 73

Passenger cabin layout Two major geometrical parameters that specify the passenger cabin are Cabin Diameter and Cabin Length . These are in turn decided by more specific details like number of seats, seat width, seating arrangement

(number abreast),

seat pitch, aisle width and number of aisles.

We choose a circular cross section for the fuselage. The overall size must be kept small to reduce aircraft weight and drag, yet the resulting shape must provide a comfortable and flexible cabin interior which will appeal to the customer

P a g e | 74

airlines. The main decision to be taken is the number of seats abreast and the aisle arrangement. The number of seats across will fix the number of rows in the cabin and thereby the fuselage length. Design of the cabin cross section is further complicated by the need to provide different classes like first class, business class, economy class etc.

Cabin length: The total number of seats 150 is distributed as 2 seats abreast. Cabin parameters are chosen based on standards for similar airplanes. The various parameters chosen are as follows

Seat pitch

= 1.06m

Seat width

= 0.46m

Aisle width

= 0.50m

Seats abreast = 5

P a g e | 75

No .of aisles =1

Hence, the total cabin length will be

=

seat

pitch*rows = 1.06*17+additional space = 23.23m

Cabin Diameter Using the number of seats abreast, seat width, aisle width we calculate the Internal diameter of the cabin.

dfus(internal) = 0.50 × 1 + 0.46 × 5 = 2.8m

P a g e | 76

According to the standards prescribed, the structural thickness is given by

t = 0.02df + 1 inch = 0.02 × 2.8+ 0.0254 = 0.0814m

Therefore the external diameter of the fuselage is obtained as 2.8 + 0.0814× 2 = 2.96 m.

P a g e | 77

Rear Fuselage: The rear fuselage profile is chosen to provide a smooth, low drag shape which supports the tail surfaces. The lower side of the profile must provide adequate clearance for aircraft when rotation during take off. The rear fuselage should also house the auxiliary power unit(APU).Based on data collected for similar aircraft we choose the ratio Ltail/dfus as 4.

Ltail = 11.2m

Total Fuselage Length

Various parts of the fuselage are indicated below

Cockpit length = 3.95m Cabin length = 23.23m

P a g e | 78

Total

= 27.18m

Centre of gravity location: The major weight components for which we have some idea of their location are the engine, the passengers and pilot, and the baggage. Using this information, we can make a preliminary estimate of the location of the centre of gravity.

Considering the forces to be acting at middle of each part, and hence taking moment about the nose, we get the centre of gravity.

Cg

=

{190*1.975+9440*(11.61+3.95)+400*17+3952*21} (190+9440+400+3952)

= 16.95m

P a g e | 79

P a g e | 80

LIFT ESTIMATION LIFT: Component of aerodynamic force generated on aircraft perpendicular to flight direction.

P a g e | 81

Lift Coefficient (CL) • Amount of lift generated depends on: – coefficient (CL) Lift=(1/2 ρ V^2)SCl=qSCl • CL is a measure of lifting effectiveness and mainly depends upon: – compressibility effects (Mach number), viscous effects (Reynolds’ number). Generation of Lift • Aerodynamic force arises from two natural sources: – Variable pressure distribution. – Shear stress distribution. • Shear stress primarily contributes to overall drag force on aircraft. • Lift mainly due to pressure distribution, especially on main lifting surfaces, i.e. wing. • Require (relatively) low pressure on upper surface and higher pressure on lower surface. • Any shape can be made to produce lift if either cambered or inclined to flow direction. • Classical aerofoil section is optimum for high subsonic lift/drag ratio.

P a g e | 82

Pressure variations with angle of attack – Negative (nose-down) pitching moment at zero-lift – Positive lift at = 0 . – Highest pressure at LE stagnation point, lowest pressure at crest on upper surface. – Peak suction pressure on upper surface strengthens and moves forwards with increasing .  – Most lift from near LE on upper surface due to suction.

P a g e | 83

Lift Curves of Cambered and Symmetrical airfoils

P a g e | 84

CALCULATION: General Lift equation is given by, Lift=(1/2 ρ V^2)SCl=qSCl Lift at Cruise 𝜌 = 0.3715 (at the cruising altitude of 10800m) V = 242.2 m/s S = 400.72 kg/m2 CL(cruise) = 0.63022 (from the wing and airfoil estimation) Substituting all these values in the general lift equation, L = 12×0.3715×242.22×400.72×0.63022 Lift at cruise = 2751761.6 N Lift at Take-Off 𝜌 = 1.225 (at sea altitude) V = 0.7 x Vlo = 0.7 x 1.2 x Vstall S = 400.72 kg/m2 CL(take-off) = 2.508 (flaps extended and kept at the take-off position of 20o) Substituting all these values in the general lift equation,

P a g e | 85

L=12×1.225×(0.7×1.2×66.86)^2×400.72×2.5

Lift at Landing 𝜌 = 1.225 (at sea altitude) V = 0.7 x Vt = 0.7 x 1.3 x Vstall S = 400.72 kg/m2 CL(landing) = 3.058 (flaps extended and kept at the landing position of 40)

Substituting all these values in the general lift equation, L =12×1.225×(0.7×1.3×60.55)^2×400.72×3.5 Lift at landing =𝟐𝟐𝟕𝟖𝟕𝟒𝟒.𝟕 𝐍

P a g e | 86

DRAG ESTIMATION DRAG: - Drag is the resolved component of the complete aerodynamic force which is parallel to the flight direction (or relative oncoming airflow). - It always acts to oppose the direction of motion. - It is the undesirable component of the aerodynamic force while lift is the desirable component.

Drag Coefficient (CD) Amount of drag generated depends on:

P a g e | 87

o Planform area (S), air density , flight speed (V), drag coefficient (CD) CD is a measure of aerodynamic efficiency and mainly depends upon: o Section shape, planform geometry, angle of attack , compressibility effects (Mach number), viscous effects (Reynolds’ number).

Drag Components - Skin Friction: o Due to shear stresses produced in boundary layer. o Significantly more for turbulent than laminar types of boundary layers



P a g e | 88

Form (Pressure) Drag o Due to static pressure distribution around body component resolved in direction of motion. o Sometimes considered separately as forebody and rear (base) drag components.

 Wave Drag o Due to the presence of shock waves at transonic and supersonic speeds.

P a g e | 89

o Result of both direct shock losses and the influence of shock waves on the boundary layer. o Often decomposed into portions related to: Lift. Thickness or Volume.

P a g e | 90

P a g e | 91

Typical streamlining effect

P a g e | 92

Lift induced (or) trailing vortex drag

The lift induced drag is the component which has to be included to account for the 3-D nature of the flow (finite span) and generation of wing lift

P a g e | 93

The lift induced drag is the component which has to be included to account for the 3-D nature of the flow (finite span) and generation of wing lift

WETTED AREA CALCULATION Wetted Area = 2*Snet {1+0.25(t/c)r 1+τλ } 1+λ

P a g e | 94

Where τ = (t/c) t / (t/c) r



(r – root, t – tip)

WING WETTED AREA: Wing wetted area

= 2 *80.1 {1+0.25(0.21) 1+

(0.95) (0.36) } 1+0.36 = 168.52m2

S (wing)wet 

TAIL WETTED AREA:

Tail Wetted Area

= 2 x 20.81 {1+0.25(0.21) 1+

(0.95) (0.29)}

1+0.29 S(tail)wet



FUSELAGE:

= 43.788m2

P a g e | 95

Fuselage wetted area

=Π x Diameter x Length

=Π (2.87) (30) =270.49m2

S(fuselage)wet

 PARASITE DRAG ESTIMATION(CDO): CDO =

Σ K Cf Swet Sref

Where, K

-

Form factor

Cf

-

Co-efficient of Friction

Sref -

Reference Wing Area

 FORM FACTOR(K): Form factor is induced to estimate the pressure drag caused due to viscous separation. It is estimated for the wing, tail and fuselage sections separately.

P a g e | 96

WING & TAIL

Kwing

=

[1+ 0.6 (0.21) + 100 (0.21)4][1.34x0.740.18 (cos

15)0.28] 0.45

Ktail

=

1.8535

=

[1+ 0.6 (0.21) + 100 (0.21)4][1.34x0.740.18 (cos

45)0.28] 0.45

=

1.6989

P a g e | 97

FUSELAGE

f (Fineness Ratio)

Kfuse

=

=

l/d

=

30 / 2.87

=

10.45

(1 + 60 10.6333

=

1.076

+ 10.633 ) 400

P a g e | 98

FORMULAS USED IN DRAG ESTIMATION

 Reynolds Number (at Altitude 40000ft) =

ρVD μ

 Co-efficient of friction (Cf)

 Total Parasite Drag Co-efficient CDO =

Σ K Cf Swet Sref

 Drag Correction for 4% Interference effect 1.04xCDO

TOTAL DRAG

CD

=

CDO + CDI

=

P a g e | 99

Induced Drag (CDI)

CL2

=

ΠeAR

CALCULATIONS: Reynolds Mach No

Cf

CDO

no.

CDO

CL

INTERFERENCE CORRECTION(4%)

700000

0.3 0.004747 0.044145

0.045910773 0.212164 0.0

1000000

0.4 0.004412 0.041028

0.042669057 0.119349 0.0

3000000

0.5 0.003595 0.033433

0.03477016 0.076379 0.0

6000000

0.6

9000000

0.7 0.002937 0.027307

0.00317 0.029475

0.030653716 0.053044 0.0 0.028398934

0.03897 5.

P a g e | 100

P a g e | 101

LANDING GEAR CONFIGURATIONS

Wheel diameter(inch) Wheel width(inch)

A

B

1.63

0.315

0.1043

0.48

MAIN LANDING GEAR No. of wheels = 4

Wheel diameter

= A (Wm) ^B

= 1.63(18,426)^0.315

P a g e | 102

= 0.91m (35.95 inch)

Wheel width

= A(Wm)^B

= 0.1043(18,426)^0.48

= 0.29m (11.63inch)

NOSE LANDING GEAR

No. of wheels = 2 Wheel diameter

= A (Wn) ^B

= 1.63(1652.43)^0.315

= 0.4272m (16.8inch)

P a g e | 103

Wheel width

= A (Wn) ^B = 0.1043(1652.43)^0.48 = 0.0927m (3.65inch)

These calculated values for diameter and width should be increased about 30% if the aircraft is to rough unpaved runways.

Determination of position of landing gear

Xacwb = xn-Vht at/a

Where,

xacwb

= aerodynamic centre of wing

body

xn

= neutral point

P a g e | 104

Vht = horizontal tail volume ratio

at/a = lift slope ratio of tail to wing

Static margin = (xn – c.g)/chord

Lets assume static margin

xn

= 18%

= 3.16*0.18+16.95 = 17.51m

From the above result, it shows that position of the landing gear is longitudinally stable.

By assuming that xacwb = x (ac)wing x (ac)wing = xn - vht

P a g e | 105

= 16.81m Therefore wing is placed such that the aerodynamic centre of the wing is placed at 16.81m behind the nose

To find the distance of leading edge of the wing from nose = 16.81 – 0.47 – 0.79 = 15.55m

P a g e | 106

Main landing gear is placed at the centre of the wing Therefore, The location of the centre of the wing = distance of the leading edge of the wing from nose + (root chord/2) = 15.55 + 4.33/2 = 17.715 m

This shows the main landing gear is located 17.71m behind the nose of the airplane. Let us locate the nose wheel so that

P a g e | 107

it can be conveniently folded rearward and upward into the fuselage. Set the nose wheel location of 4.43m as shown.

3-D VIEW DIAGRAM

P a g e | 108

P a g e | 109

AIRCRAFT PERFORMANCE THRUST REQUIRED:

The aircraft to be designed is assumed to be in a steady, level flight at an altitude of 30,000ft and at the given cruise velocity. The aircraft’s power plant must produce the net thrust to overcome the drag, in a sense that the net thrust produced is equal to the drag experienced by the aircraft. The thrust required to obtain this steady velocity is given by the equation, Thrust Required,

TR

=

_W_ CL/CD

FORMULAE For a given value of velocity (V),  CL

=

2W ρV2S

 CD =

CDO +

CL2 ΠeAR

 From the ratio (CL/ CD) and the weight of the aircraft (W), the thrust required is estimated.

P a g e | 110

P a g e | 111

RATE OF CLIMB

The aircraft is considered to be in steady, unaccelerated, climbing flight.

The rate of climb is calculated from the excess Power denoted in the Fig.

R/C =

Excess Power W

=

5008.5 x 103

P a g e | 112

34654.91 x 9.81 R/C =

14.73 m/s

or

883.9 m/min

RANGE & ENDURANCE Range is technically defined as the total distance traversed by the aircraft on a tank of fuel. Endurance is the total time the aircraft stays on air on a tank of fuel. One of the critical parameters influencing range and endurance is the Thrust Specific Fuel Consumption which amount of fuel consumed per unit thrust per unit time. Range and Endurance are found using the Brequet Formula. ENDURANCE E

=

1 CL ln WO Ct CD

Where, Ct

`

W1

- Thrust Specific Fuel Consumption

WO - Gross Weight of aircraft W1

- Weight of aircraft without Fuel

P a g e | 113

E

=

___1

x 1.36 x ln 34654.91

1.9425x10-5

34654.91 -

7894.22 = 301 min or 5.02 hrs RANGE R

R

=

=

Range =

2

_2 CL1/2 (WO1/2 - W11/2)

Ct

ρ S CD

2 x 0.2334 x 0.514x105 x 6.937 x 22.57

3756.6 km

TAKEOFF PERFORMANCE Up to this point we have discussed the aircraft performance at zero accelerations. But finite accelerations are required for

P a g e | 114

further estimation of parameters. For takeoff the pilot accelerates with afterburning thrust of 60kN.

Takeoff Distance

=

1.44 W2 g ρ S CL,max T

=

1.44 x (34654.91 x 9.81)2 9.81 x 1.225 x 80.1 x 2 x 60000

Takeoff Distance

=

1440.82 m

LANDING PERFORMANCE The ground roll after the plane has touched down has to be calculated. To minimize the distance required for complete stop the pilot has decreased the thrust to zero after touched down, so T=0. Spoilers are deployed so lift tends to zero. Drag shoot is also actuated leads to 20%

P a g e | 115

increase in drag. The maximum lift coefficient with flaps fully employed at touchdown is 2.4.

Landing Distance(SL)

=

1.69 W2

g ρ S CL,max {D + μr(W-L)} where, μr

SL

-

Co-efficient of rolling friction (0.4)

= _________1.69(34654.91 x

9.81)2__________________ 9.81 x 1.225 x 80.1 x 2.4 x (23101.5 + 0.4(34654.91 x 9.81))

Landing Distance

=

531.45 m

P a g e | 116

P a g e | 117

BIBLIOGRAPHY

REFERENCES: 1. Airplane Design by Dr. Jan Roskam, 3rd edition. 2. Aircraft Design: A Conceptual Approach by Daniel P. Raymer, 4th edition. 3. Introduction to Flight by John D. Anderson, 2nd edition. 4. Aircraft Performance and Design by John D. Anderson, 2nd edition. 5. Theory of wing sections by Ira.H.Abbott, Dover edition.

P a g e | 118

View more...

Comments

Copyright ©2017 KUPDF Inc.
SUPPORT KUPDF