AIRCRAFT DESIGN PROJECT-1150 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. . To assess the development potential in the primary role of the aircraft. To produce a commercial analysis of the aircraft project.7|Page To offer a unique and competitive service to existing scheduled operations. 8|Page ABSTRACT The purpose of the project is to design a 150 seater Medium Range International passenger aircraft. It must possess turbofan engines to provide the required amount of speed. tricycle landing gear and a conventional tail arrangement. Such an aircraft must possess a wide body configuration to provide sufficient seating capacity. range and fuel economy for the operator. . The aircraft will possess a low wing. The aircraft will possess two engines. Airbus started the development of a very large airliner (termed Megaliner by Airbus in the early development stages) in the early 1990s. This report gives the different aspects of specifications like wing specification. we intend to implant the differentiation among the aircrafts having sitting capacity of 100-180 members. Every year there is a great competition for making an aircraft of having higher capacity of members inside the aircraft. McDonnell Douglas pursued a similar strategy with its ultimately unsuccessful MD-12 design.9|Page INTRODUCTION At the instant time there are different types of aircrafts with latest technology. they knew there was room for . As each manufacturer looked to build a successor to the 747. weight specification. So here in this report. 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. power plant specification and performance specification. Boeing and several companies in the Airbus consortium started a joint feasibility study of an aircraft known as the Very Large Commercial Transport (VLCT). it must be able to lift the weight of the airplane. . landing gear. its fuel. For any airplane to fly. which are usually located beneath the wings. eventually resulting in Lockheed's departure from the civil airliner business.10 | P a g e only one new aircraft to be profitable in the 600 to 800 seat market segment. the airplane must be pushed through the air. but all modern airplanes have certain components in common. In January 1993. These are the fuselage. Each knew the risk of splitting such a niche market. Airplanes come in many different shapes and sizes depending on the mission of the aircraft. To generate lift. The engines. 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. and powerplant. the passengers. The wings generate most of the lift to hold the plane in the air. aiming to form a partnership to share the limited market. tail assembly and control surfaces. wing. provide the thrust to push the airplane forward through the air. and the cargo. . The fuselage is generally streamlined as much as possible to reduce drag. 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. In addition.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 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). an engine may be housed in the fuselage. depend upon the airplane mission and the best compromise necessary in the overall airplane design. as well as the airfoil section shape. The wing provides the principal lifting force of an airplane. Some aircraft carry fuel in the fuselage. 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. others carry the fuel in the wings. 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. .12 | P a g e The control surfaces include all those moving surfaces of an airplane used for attitude. lift. They include the tail assembly. and drag control. 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 .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. travel Preliminary and Structural Testing Drafting the preliminary 3-view Drawings . 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. This Brochure has to be APPROVED by the manufacturer and/or the customer.14 | P a g e DESIGN CYCLE PRELIMINARY DESIGN It consists of the initial stages of design.g. resulting in the presentation of a BROCHURE containing preliminary drawings and clearly stating the operational capabilities of the airplane being designed. 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. limits Final performance calculation . 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. which will consist of Construction of Mock-up Structural layout of all the individual units. 16 | P a g e CYCLES OF DESIGN PROCESS: . .17 | P a g e Aircraft design can be broken into three major phases. and other changes. Design requirements include aircraft range and payload. New ideas and problems emerge as a design is investigated in ever increasing detail. it must be redrawn to reflect the new gross weight. fuel weight. 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. wing size. Each time the latest design is analyzed and sized. and maneuverability and speed requirements.18 | P a g e (a) Conceptual Design (b) Preliminary Design (c) Detail Design CONCEPTUAL DESIGN: Conceptual design is a very fluid process. . engine size. takeoff and landing distances. and the internal locations of the major components such as the engine. the fuselage shape. landing gear and fuel tanks. The big questions such as whether to use a canard or an aft tail have been resolved. cockpit. At some point late in preliminary design. A good conceptual sketch will include the approximate wing and tail geometries. PRELIMINARY DESIGN: It can be said to begin when the major changes are over. even minor changes are stopped when a decision is made to freeze the . payload/passenger compartment.19 | P a g e The actual design effort usually begins with conceptual sketch. also called “FULL-SCALE DEVELOPMENT”. The ultimate objective during this design is to ready the company for the detail stage. Lofting is the mathematical modeling of the outside skin of the aircraft with sufficient accuracy to insure proper fit between its different parts. structures. and stability and control. During this design the specialists in areas such as structures. propulsion. A key activity during this type of design is “LOFTING’. and control systems will design and analyze their portion of the aircraft. even if they are designed by different designers and possibly fabricated in different locations. DETAIL DESIGN: . Testing is initiated in areas such as aerodynamics. landing gear.20 | P a g e configuration. during conceptual and preliminary design the wing box will be designed and analyzed as a whole. the detail design phase begins in which the actual pieces to be fabricated are designed. During detail design. starting with smallest and simplest subassemblies and building upto the final assembly process. but the design must still meet the original requirements. . each of which must be separately designed and analyzed. Specialists determine how the airplane will be fabricated. spars. Another important part of detail design is called production design. Production designers frequently wish to modify the design for ease of manufacture. For example. and skins. that whole will be broken down into individual ribs.21 | P a g e Assuming a favorable decision for entering full-scale development. Compromises are inevitable. that can have a major impact on performance or weight. Control laws for the flight control system are tested on an “iron-bird” simulator.22 | P a g e During detail design. the testing effort intensifies. Actual structure of the aircraft is fabricated and tested. a detailed working model of the actuators and flight control surfaces. Comparative Datasheet – 1 . 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. Flight simulators are developed and flown by both company and customer test pilots. 35 28.35 HEIGHT (M) 11.53 37.77 28.5 ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) POWER PLANT 0.23 WING AREA(m^2) 102 102 THRUST (kN) 64 77 EMPTYWEIGHT(kg) 28100 MAX TAKE 121 33.55 57150 OFF 50300 WEIGHT 136 117 30.2 12.867 P&T JT8D RR BR715A1 .12 P&T JT8D Cfm56-7 Comparative Datasheet – 2 786 817.407 786 839.23 8.3 34.87 82.3 31600 52400 49900 10700 2645 78000 SERVICE SEILING (m) 10670 RANGE (km) 2850 ASPECT RATIO 8 12500 5600 4300 8 8.23 | P a g e NAME OF AIRCRAFT Boeing Boeing C-40 Boeing Boeing 737-100 Clipper 717-200 737-200 CAPACITY 124 LENGTH (M) 28.48 11.65 WING SPAN (M) 28. 11 9.8 HEIGHT(M) 12.6 THRUST( kN) TAKE 106 WEIGHT (Kg) RANGE 5700 4200 4444 5970 ASPECT RATIO 10 9.414 31.53 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) 112.1 28.488 POWER PLANT CMF56-5 CMF56-7 .8 ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 828.15 11.44 33.9 35.51 11.008 31.9 28.46 9.24 | 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) 31.45 786 786 833.2 WING SPAN(M) 34.15 12. 92 9.8 34 36.25 | 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) 33.8 POWER PLANT CMF56-7 RR BR715 GE CF34 .55 8.47 38 36.5 121 80 WING AREA (M^2) THRUST (kN) 117 82.63 37.7 ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 833.36 WING SPAN (M) 35.35 HEIGHT (M) 12.3 EMPTY WEIGHT (Kg) 38147 30618 MAX TAKE OFF 66000 49900 82.3 827.8 8.44 84.3 65700 26300 43616 WEIGHT (Kg) SERVICE SEILING (M) 12500 11000 11000 11900 RANGE (kM) 6370 2645 2532 2200 ASPECT RATIO 8 7.9 817.9 734.8 7 7.8 28. 53 35.2 61.65 12.08 39.5 8.5 93.5 12.9 HEIGHT (M) 8.53 41.5 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 .25 44.08 28.26 | 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) 35.07 WING SPAN (M) 28.5 THRUST (kN) 67.9 39.93 WING AREA (M^2) 93.6 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. 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 6.3 8.7 28.9 862 862 798 ENDURANCE WING LOADING THRUST TO WEIGHT RATIO CRUISESPEED(km/h) 968.32 THRUST (kN) 67.9 32.61 40.93 10.M) Antonov An-158 87.91 HEIGHT (M) 12.6 WING ARE (Sq.4 POWER PLANT PW JTD-3D PW JT8D-7 D-36 .9 28.27 | P a g e NAME OF AIRCRAFT Boeing Boeing Boeing 707-320B 727–100 727–200 CAPACITY 147 149 189 99 LENGTH (M) 46.3 10.6 46.91 WING SPAN (M) 44.42 32. 28 | P a g e COMPARITIVE GRAPHS: . 29 | P a g e LENGTH: 3o RANGE:4800KM . 3 .30 | P a g e ASPECT RATIO:9. 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 . M .36 | P a g e WING LOADING:547 Kg/Sq. 37 | P a g e WING SPAN:27.2M . 38 | P a g e WING AREA:80 M^2 . 2 m (89ft 2 inch) WING AREA (S) 80.39 | P a g e DESIGN DATA SHEET GENERAL CHARACTERISTICS CREW 4 PASSENGER CAPACITY 100-180 LENGTH 30 m (98ft 5 inch) WING SPAN (b) 27.18 ft2) ASPECT RATIO(b2/S) 9.1 m2 (862.405 kg (22.939 lb) POWERPLANT ROLLS ROYCE TAY 620-15 TURBOFAN NO OF ENGINES 2 .710 lb) FUEL WEIGHT 10.506 kg (84.3 MAX TAKE OFF WEIGHT 38.891 lb)(corrected) EMPTY WEIGHT 31620 kg (69. 488 lbf) PERFORMANCE CHARACTERISTICS MAX SPEED 940 km/h (0.640 m(38.78 mach. The reserve fuel as mentioned above is needed during loiter.995 miles) SERVICE CEILING 11. 584 mph) CRUISE SPEED 890 km/h (0.40 | P a g e THRUST 60 KN (13. .74 mach.820 km (n 2.553 mph) RANGE 4. The need of fuel is necessary only for the shown cruise.188 ft) MISSION SPECIFICATION The aircraft undergoes a simple mission. followed by 1 hour loiter followed by a 100nm flight to alternate. CRUISE SPEED: M=0. POWERPLANTS: 2 Turbo-fan engines. CREW: 2 pilots and 2 cabin attendants at 200lbs and 30lbs baggage each respectively. WTO is desired.100 ft.100 feet (For the design range).41 | P a g e PAYLOAD: 150 passengers at 75kg each and 25kg of baggage each. The need of fuel is necessary only for the shown cruise. The reserve fuel as mentioned above is needed during loiter. CLIMB: A dir6-ect climb to 33. ALTITUDE: 33. at 38. MISSION SPECIFICATION The aircraft undergoes a simple mission.74. at Max. . RANGE: 2502nm.100 ft. CRUISE SPEED: M=0.100 feet (For the design range). WTO is desired. followed by 1 hour loiter followed by a 100nm flight to alternate.100 ft.42 | P a g e PAYLOAD: 150 passengers at 75kg each and 25kg of baggage each. at Max. . at 38. CREW: 2 pilots and 2 cabin attendants at 75kgs and 25kgs baggage each respectively.100 ft.74. RANGE: 2502nm. ALTITUDE: 33. POWERPLANTS: 2 Turbo-fan engines. CLIMB: A direct climb to 33. 43 | P a g e . We know the lift acquired is directly proportional to the weight of the aircraft.44 | P a g e WEIGHT ESTIMATION Airplane must normally meet very stringent range. Hence the estimation of weight for a given aircraft plays a key role in design analysis. 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. endurance. W= Wstructure + Wpayload + Wfuel + Wcrew +Wpower plant + Wfixed equipment . +Baggage) 4*(75+10) = 340 kg Approximate Take-off Weight(WTO(approx)): .of baggage) = WPL = 150*(75+25) 15000 kg Crew Weight Validation (WCREW): WCrew = WCrew = No. of crew members*(Crew wt.of passenger+Wt. of passengers*(Wt.45 | P a g e Payload Weight Validation (WPL): WPL = No. 46 | P a g e WTO(approx)= 38393 kg [From Design data sheet] Now. taking into consideration the appropriate mission phases: . 995 0.985 1.992 0.995 0.985 0.980 0.995 0.992 For takeoff.97.87 0.990 0.000 0.980 0.990 0.995 0. segment 0-1 historical data’s shows that.970 0. . ?1/?0 = 0.990 0.P a g e | 47 Airplane Take Off Climb Type Business Jets Transpor t Military Trainers Superson ic Cruise Descent Landing 0.92-0. ?5/?4 = 0. segment 1-2 historical data shows that.1 For loiter.995 For landing. ?5/?4 = 0. segment 3-4 ignoring the fuel consumption during descent we assume.995 .P a g e | 48 For climb.985 W1 Aircraft Design Project . W2 = 0. ?4/?3 = 1 For landing. segment 4-5 based on historical data we assume that. segment 4-5 based on historical data we assume that. ?/? =15 From the comparative data sheet.P a g e | 49 The Brequet’s range equation is used to calculate the value of ?3?2.6 hr-1 . R = (?∞/??)(?/?)ln (?2/?3) 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. As we all know that maximum range is covered during cruise we considering this equation. V∞ = 872 km/hr R = 7200 km For Business & transport jets We found the values of cj as 0. So. 8x1x0.25 W3/w2=0.6)(15)ln(W2/W3) ln(w2/w3)=(4820/23500)= 0.P a g e | 50 So now substituting these values in the Brequet’s range equation. R = (?∞/(?/?)(ln ?2/?3) = (940/.995 .97x0.8 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.225 (w2/w3)=1.985x0. MFF = (w1/w0)x(w2/w1)x(w3/w2)x(w4/w3)x(w5/w4) = 0. 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 114 Wg Hence WPP = 3921.15 20811 kg .P a g e | 54 Hence WFE = 5828 kg WPP = 0.54+5828.5 kg Hence the empty weight WE = WE = = Wstruc +WPP + WFE 11061.45+3921. A consolidated list of the various engines that tally with the thrust required for the given aircraft is tabulated below.P a g e | 55 POWERPLANT SELECTION As estimated from the design data sheet. the aircraft to be designed requires a thrust of 60kN.hr) Max Thrust (kN) . Engine Dry Name weight SFC (Kg/kN. 02 54.P a g e | 56 (kg) RollsRoyce 1501 69.00 TAY 620 GE-CF348 RollsRoyce TAY 650 RollsRoyce BR710-48 RollsRoyce .60 1200 71.16 2104 72.93 61.54 1595 69.2 64.97 66.93 67.72 1856 77. namely Rolls-Royce TAY 620 Turbo Fan Engine to meet the given design standards.P a g e | 57 Hence. the optimum choice of engine. Rolls-Royce TAY 620 Description . we select 2 rear engines. In our design. 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. 4 m(94.hr) (SFC) Fuel consumption(Eg 300 NM 2063 KG flight) .17 m(44 inch) Engine weight 14220.04 Overall Pressure Ratio 16.hr(0.93Kg/kN.0 Fan diameter 1.8 inch) Turbine entry Temperature 1305 K Specific fuel consumption 69.69 lb/lbf.P a g e | 58 Manufacturer ROLLS-ROYCE Type of the engine TAY -620 TURBOFAN Thrust at SL/ISA 61.57 Kg(3135lbs) Engine length 2.6 kN (13850 lbs) Inlet mass flow 184 Kg/s(408 lb/s) Bypass Ratio 3. P a g e | 59 ROLLS-ROYCE TAY620 . 303 Cruise velocity = 872km/hr = 242. it is now possible to estimate the amount of fuel required for a flight at the given cruising speed for the given range.3031.P a g e | 60 FUEL WEIGHT VALIDATION The choice of a suitable engine.3715/1.?)/ ?????? ???????? The factor of 1. Wfuel = (?????? ?? ???????∗?????? ?? ????????∗?????∗???∗?. Thrust at altitude is calculated using the relation: =Ρalt/ ρo Altitude = 10800m = 35433ft ? = ? ??? ?0 = 0.225 = 0.2kg SFC = 0.2 ??= 76.2 is provided for reserve fuel.363kN = 7784. having been made.4hr-1 (at medium thrust setting) .2m/s To = 320kN ??= 320×0. 2 872 Wfuel = 92.4×1. Determination of mean aerodynamic chord 3.553. we have two considerations: 1. The geometric shape of the wing 2. Wing location relative to the fuselage .2×7200×0.42 kg WING SELECTION AERODYNAMICS Wing Configuration: In the wing design.P a g e | 61 Number of engines = 3 CALCULATION: Wfuel = 3×7784. The maximum design velocity is 940 km/h (.78 M) . Aerofoil shape and thickness along the span f. We choose to use a swept wing.below the transonic region. Taper ratio e. The parameters known include: . Aspect ratio (which is already obtained from comparative graphs) c. Wing sweep d.P a g e | 62 GEOMETRY OF THE WING: Wing geometry is described by a. Geometric twist (change in aerofoil chord incidence angle along the span). Plan-form shape b. Initially the primary aerodynamics data was obtained from the design data sheet. 1 m^2 Aspect ratio.2m Wing Area.3 Taper Ratio λ = 0. b = 27. λ = Ct = 0. AR = b^2/S = 9.36Cr Where. S = 80. Ct is the root chord Cr is the tip chord Ct =0.2/2 ((0.P a g e | 63 Wing span.36 Cr Also S/2 = b/2 (Ct + Cr )/2 80.1/2 = 27.36 Cr + Cr ) / 2) .36 (from Roskam book – BAE 196-200) We know that taper ratio. Hence we get the plan form as shown .P a g e | 64 Hence Cr = 4.56m A leading edge sweep angle of 15°.33m Also Ct = 1. 16 m Distance of the mean chord = {b*(1+2 λ)}/{6*(1+λ)} from the aircraft centre line = 5.P a g e | 65 DETERMINATION OF THE MEAN AERODYNAMIC CHORD: Mean chord = {(2/3)*Cr (1+λ+ λ2)}/ (1+λ) = 3.73 m . Low-slung fuselage – ease to place the fuselage lower to the ground. Low wing High wing: 1. Mid wing 3. 3.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. . 2. It is a distinct advantage for transport plane since it simplifies the loading and unloading processes. More stable in lateral and rolling motion. 2. Gives best stability with little dihedral. Landing gear can easily be retracted into the wing box. 2. SELECTION OF AEROFOIL: .P a g e | 67 Mid wing: 1. Added fillet will avoid undesirable aerodynamic interference. Least interference drag. In light of all the above considerations. Low wing: 1. we choose a highwing configuration mainly due to structural and Landing gear considerations. 965 m^3 . 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)}*0.33 Volume of fuel = 2*{((2/3)*(t/c)*c)*c*(b/4)}*2 Volume of fuel = Weight of fuel(kg) Specific gravity of fuel = 7894.5*1.P a g e | 68 In order to select an aerofoil the appropriate thickness to chord ratio(t/c) is to be determined.716 (0.72*1000) = 10. 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 10.5*1.21 As our aircraft will fly only at subsonic speeds.2/4)}*0.945)*2. Based on the t/c ratio.33 t/c = 0.945*(27.209 (or) 0. we have chosen a NACA 6 series aerofoil. . we have chosen the aerofoil NACA 664-221.965 = 2*{((2/3)*(t/c)*2. The fuselage of a passenger aircraft is divided into a number of a sections: 1. Cockpit 3.P a g e | 70 FUSELAGE LAYOUT The fuselage layout is important as the length of the entire aircraft depends on this. Cabin 4. The length and diameter of the fuselage is related to the seating arrangement. Tail fuselage . Nose 2. has a major impact upon the overall shape. Overall effect depends on the level of pressurization.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 sizing: . fuselage layout proceeds almost independent of other design aspects Pressurisation If required. b=0.a=0. flight decks of various airplanes are considered and the value of Lnose/Lfus is found to be 0.58m .5 = 0.P a g e | 72 A relation between the gross weight of the aircraft and the length of the aircraft.67. For jet transport.03 Lnose = 0. Lfus = aWb Where W is in lbs and Lfus is in ft.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.03*19.43 Lfus = 19. 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. The overall size must be kept small to reduce aircraft weight and drag. We choose a circular cross section for the fuselage. seat width. aisle width and number of aisles. seat pitch. seating arrangement (number abreast). yet the resulting shape must provide a comfortable and flexible cabin interior which will appeal to the customer . Cabin length: The total number of seats 150 is distributed as 2 seats abreast.46m Aisle width = 0. The main decision to be taken is the number of seats abreast and the aisle arrangement. economy class etc.06m Seat width = 0.P a g e | 74 airlines.50m Seats abreast = 5 . Cabin parameters are chosen based on standards for similar airplanes. The various parameters chosen are as follows Seat pitch = 1. Design of the cabin cross section is further complicated by the need to provide different classes like first class. business class. The number of seats across will fix the number of rows in the cabin and thereby the fuselage length. the total cabin length will be = seat pitch*rows = 1. seat width. aisle width we calculate the Internal diameter of the cabin.P a g e | 75 No . dfus(internal) = 0.50 × 1 + 0.23m Cabin Diameter Using the number of seats abreast.8m .of aisles =1 Hence.46 × 5 = 2.06*17+additional space = 23. 96 m.0254 = 0.02 × 2.0814× 2 = 2.0814m Therefore the external diameter of the fuselage is obtained as 2.P a g e | 76 According to the standards prescribed.02df + 1 inch = 0. .8 + 0. the structural thickness is given by t = 0.8+ 0. The rear fuselage should also house the auxiliary power unit(APU).2m Total Fuselage Length Various parts of the fuselage are indicated below Cockpit length = 3. Ltail = 11.95m Cabin length = 23.23m .Based on data collected for similar aircraft we choose the ratio Ltail/dfus as 4.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. Using this information. and the baggage. and hence taking moment about the nose.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. Considering the forces to be acting at middle of each part.61+3. we can make a preliminary estimate of the location of the centre of gravity. we get the centre of gravity.95)+400*17+3952*21} (190+9440+400+3952) = 16. Cg = {190*1.95m .975+9440*(11. the passengers and pilot. P a g e | 79 . P a g e | 80 LIFT ESTIMATION LIFT: Component of aerodynamic force generated on aircraft perpendicular to flight direction. . • Lift mainly due to pressure distribution.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: – Section shape. wing. • Shear stress primarily contributes to overall drag force on aircraft. i. • Require (relatively) low pressure on upper surface and higher pressure on lower surface. – Shear stress distribution. especially on main lifting surfaces. Generation of Lift • Aerodynamic force arises from two natural sources: – Variable pressure distribution.e. . • Classical aerofoil section is optimum for high subsonic lift/drag ratio. viscous effects (Reynolds’ number). • Any shape can be made to produce lift if either cambered or inclined to flow direction. compressibility effects (Mach number). 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. – Peak suction pressure on upper surface strengthens and moves forwards with increasing . lowest pressure at crest on upper surface. – Most lift from near LE on upper surface due to suction. . P a g e | 83 Lift Curves of Cambered and Symmetrical airfoils . 63022 Lift at cruise = 2751761.3715 (at the cruising altitude of 10800m) V = 242.2 m/s S = 400.P a g e | 84 CALCULATION: General Lift equation is given by.7 x Vlo = 0.6 N Lift at Take-Off ? = 1.72 kg/m2 CL(cruise) = 0.72×0. .3715×242.7 x 1.72 kg/m2 CL(take-off) = 2. L = 12×0.63022 (from the wing and airfoil estimation) Substituting all these values in the general lift equation. Lift=(1/2 ρ V^2)SCl=qSCl Lift at Cruise ? = 0.225 (at sea altitude) V = 0.2 x Vstall S = 400.22×400.508 (flaps extended and kept at the take-off position of 20o) Substituting all these values in the general lift equation. 225×(0.86)^2×400.7 x Vt = 0. L =12×1.3×60.058 (flaps extended and kept at the landing position of 40) Substituting all these values in the general lift equation.7×1.P a g e | 85 L=12×1.3 x Vstall S = 400.225×(0.5 Lift at Landing ? = 1.72×2.? ? .7×1.225 (at sea altitude) V = 0.72×3.5 Lift at landing =???????.72 kg/m2 CL(landing) = 3.7 x 1.55)^2×400.2×66. Drag is the resolved component of the complete aerodynamic force which is parallel to the flight direction (or relative oncoming airflow). . .P a g e | 86 DRAG ESTIMATION DRAG: .It is the undesirable component of the aerodynamic force while lift is the desirable component.It always acts to oppose the direction of motion. Drag Coefficient (CD) Amount of drag generated depends on: . drag coefficient (CD) CD is a measure of aerodynamic efficiency and mainly depends upon: o Section shape. planform geometry. angle of attack . air density . flight speed (V). viscous effects (Reynolds’ number).Skin Friction: o Due to shear stresses produced in boundary layer. o Significantly more for turbulent than laminar types of boundary layers . Drag Components . compressibility effects (Mach number).P a g e | 87 o Planform area (S). . 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 | 88 Form (Pressure) Drag o Due to static pressure distribution around body component resolved in direction of motion. o Often decomposed into portions related to: Lift.P a g e | 89 o Result of both direct shock losses and the influence of shock waves on the boundary layer. 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 . 95) (0.95) (0. t – tip) WING WETTED AREA: Wing wetted area = 2 *80.21) 1+ (0.21) 1+ (0.29 S(tail)wet FUSELAGE: = 43.25(0.788m2 .36) } 1+0.1 {1+0.52m2 S (wing)wet TAIL WETTED AREA: Tail Wetted Area = 2 x 20.25(0.29)} 1+0.P a g e | 94 Where τ = (t/c) t / (t/c) r (r – root.81 {1+0.36 = 168. 49m2 S(fuselage)wet PARASITE DRAG ESTIMATION(CDO): CDO = Σ K Cf Swet Sref Where. It is estimated for the wing. tail and fuselage sections separately.P a g e | 95 Fuselage wetted area =Π x Diameter x Length =Π (2.87) (30) =270. . 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. 28] 0.18 (cos 45)0.34x0.45 = 1.45 Ktail = 1.18 (cos 15)0.28] 0.8535 = [1+ 0.21)4][1.21)4][1.6 (0.740.21) + 100 (0.21) + 100 (0.6989 .740.P a g e | 96 WING & TAIL Kwing = [1+ 0.34x0.6 (0. 87 = 10.633 ) 400 .P a g e | 97 FUSELAGE f (Fineness Ratio) Kfuse = = l/d = 30 / 2.45 (1 + 60 10.6333 = 1.076 + 10. 04xCDO TOTAL DRAG CD = CDO + CDI = .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. 002937 0.0 6000000 0. .045910773 0.0 3000000 0.00317 0.03477016 0.028398934 0.029475 0.4 0.7 0.004412 0.119349 0.P a g e | 99 Induced Drag (CDI) CL2 = ΠeAR CALCULATIONS: Reynolds Mach No Cf CDO no.03897 5.033433 0.076379 0.003595 0.212164 0.030653716 0.004747 0.053044 0.3 0.6 9000000 0.044145 0.027307 0.042669057 0.0 1000000 0.0 0.041028 0.5 0. CDO CL INTERFERENCE CORRECTION(4%) 700000 0. P a g e | 100 . 63 0.1043 0.315 0.48 MAIN LANDING GEAR No.426)^0. of wheels = 4 Wheel diameter = A (Wm) ^B = 1.63(18.P a g e | 101 LANDING GEAR CONFIGURATIONS Wheel diameter(inch) Wheel width(inch) A B 1.315 . of wheels = 2 Wheel diameter = A (Wn) ^B = 1.1043(18.48 = 0.43)^0.91m (35.8inch) .315 = 0.P a g e | 102 = 0.63inch) NOSE LANDING GEAR No.29m (11.95 inch) Wheel width = A(Wm)^B = 0.4272m (16.426)^0.63(1652. 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.0927m (3. xacwb = aerodynamic centre of wing body xn = neutral point .1043(1652.48 = 0.43)^0.P a g e | 103 Wheel width = A (Wn) ^B = 0. 18+16. it shows that position of the landing gear is longitudinally stable.P a g e | 104 Vht = horizontal tail volume ratio at/a = lift slope ratio of tail to wing Static margin = (xn – c.51m From the above result. By assuming that xacwb = x (ac)wing x (ac)wing = xn .95 = 17.g)/chord Lets assume static margin xn = 18% = 3.16*0.vht . 81m behind the nose To find the distance of leading edge of the wing from nose = 16.81m Therefore wing is placed such that the aerodynamic centre of the wing is placed at 16.79 = 15.55m .47 – 0.P a g e | 105 = 16.81 – 0. 33/2 = 17.715 m This shows the main landing gear is located 17.55 + 4.71m behind the nose of the airplane. Let us locate the nose wheel so that .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. 3-D VIEW DIAGRAM .P a g e | 107 it can be conveniently folded rearward and upward into the fuselage.43m as shown. Set the nose wheel location of 4. P a g e | 108 . CL = 2W ρV2S CD = CDO + CL2 ΠeAR From the ratio (CL/ CD) and the weight of the aircraft (W). The thrust required to obtain this steady velocity is given by the equation. in a sense that the net thrust produced is equal to the drag experienced by the aircraft. The aircraft’s power plant must produce the net thrust to overcome the drag. . the thrust required is estimated.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. Thrust Required.000ft and at the given cruise velocity. TR = _W_ CL/CD FORMULAE For a given value of velocity (V). P a g e | 110 . climbing flight. The rate of climb is calculated from the excess Power denoted in the Fig.P a g e | 111 RATE OF CLIMB The aircraft is considered to be in steady. unaccelerated.5 x 103 . R/C = Excess Power W = 5008. Range and Endurance are found using the Brequet Formula.Gross Weight of aircraft W1 .91 x 9.P a g e | 112 34654.73 m/s or 883.81 R/C = 14. Endurance is the total time the aircraft stays on air on a tank of fuel.9 m/min RANGE & ENDURANCE Range is technically defined as the total distance traversed by the aircraft on a tank of fuel. ENDURANCE E = 1 CL ln WO Ct CD Where. Ct ` W1 .Weight of aircraft without 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.Thrust Specific Fuel Consumption WO . 514x105 x 6.2334 x 0.91 - 7894.937 x 22.W11/2) Ct ρ S CD 2 x 0.9425x10-5 34654.91 1. But finite accelerations are required for .02 hrs RANGE R R = = Range = 2 _2 CL1/2 (WO1/2 .6 km TAKEOFF PERFORMANCE Up to this point we have discussed the aircraft performance at zero accelerations.P a g e | 113 E = ___1 x 1.57 3756.22 = 301 min or 5.36 x ln 34654. 44 W2 g ρ S CL.91 x 9.81 x 1. Spoilers are deployed so lift tends to zero. For takeoff the pilot accelerates with afterburning thrust of 60kN.225 x 80.44 x (34654.82 m LANDING PERFORMANCE The ground roll after the plane has touched down has to be calculated. so T=0.P a g e | 114 further estimation of parameters. To minimize the distance required for complete stop the pilot has decreased the thrust to zero after touched down.max T = 1. Drag shoot is also actuated leads to 20% .1 x 2 x 60000 Takeoff Distance = 1440.81)2 9. Takeoff Distance = 1. 4) = _________1.max {D + μr(W-L)} where. μr SL - Co-efficient of rolling friction (0.225 x 80.69 W2 g ρ S CL. Landing Distance(SL) = 1.4(34654.1 x 2.91 x 9.81 x 1. The maximum lift coefficient with flaps fully employed at touchdown is 2.5 + 0.P a g e | 115 increase in drag.69(34654.45 m .81)2__________________ 9.81)) Landing Distance = 531.4 x (23101.91 x 9.4. P a g e | 116 . 3rd edition. 5. 4th edition. Aircraft Design: A Conceptual Approach by Daniel P. 4. Aircraft Performance and Design by John D. 2. Jan Roskam.P a g e | 117 BIBLIOGRAPHY REFERENCES: 1. 3. Dover edition. 2nd edition. Airplane Design by Dr.H. 2nd edition.Abbott. Anderson. Theory of wing sections by Ira. Raymer. Anderson. Introduction to Flight by John D. . P a g e | 118 .