УДК 629.7Інв. №

МІНІСТЕРСТВО ОСВІТИ І НАУКИ, МОЛОДІ ТА СПОРТУ УКРАЇНИНаціональний аерокосмічний університет ім. М. Є. Жуковського

«Харківський авіаційний інститут»

Кафедра проектування літаків i вертольотів

ДО ЗАХИСТУ ДОПУСКАЮ

Завідувач кафедри

д–р техн. наук, проф.

_________О. Г. Гребеніков

(підпис)

PASSENGER AIRCRAFT INTEGRATED DESIGNING AND MODEL ANALYSIS

Пояснювальна записка до випускної роботи магістра,напрямок 8.100101 — «Авіація та космонавтика»

Фах — «Літаки і вертольоти»

(номер зал. книжки без позначки «№»)

Виконавець студент гр.   16Е- 2 KIRUBAGARAN MAZHALAI PRIYAN

(№ групи) (І.Б.П.)

(підпис, дата)

Керівник–консультант з основного розділу

к. техн. наук, доц. S. Trubaev

(науковий ступінь, вчене звання) (І.Б.П.)

(підпис, дата)

Нормоконтролер к. техн. наук, доц. S. Trubaev

(науковий ступінь, вчене звання кєрівника)(І.Б.П.)

(підпис, дата)

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Ministry of Science and Education, Youth and Sports of UkraineNational Aerospace University,named by N.E. Zhukovskiyj

«Kharkov Aviation Institute»

Faculty of aircraft and helicopter construction

Aircraft and helicopter design department

«Approved by»Head of department №103,

prof.___________ Grebenikov А. G.«___»_______________ 201__

ASSIGNMENT

FOR A FINAL WORK OF AN APPLICANT

for a masters degree 8.05110101 «Aircraft and helicopter»

group 16 E- 2       students name KIRUBAGARAN MAZHALAI (Name)

SUBJECT OF GRADUATION PROJECT

«PASSENGER AIRCRAFT INTEGRATED DESIGNING AND MODEL

ANALYSIS»

Initial data for design: Vmax – _835____ km/h; Vkr – __760__ km/h; Vу – __14___ m/s; Hmax

– _11____ km; Hcr – __10___ km; L – ___2000__ km; Lp – __1970___ m; npas. – _ 47____ men; ncrew – ___4__ men.; mp/l – _39___ t;Т – __4____h; Кcr – __0.10___.

Graduation Project. Table of ContentsAbstractDesign section1. Computer–aided general design of aircraftIntroduction, design goal – setting and tasking

1.1. Purpose, aircraft performance requirements, conditions of production and operation, limitations imposed by aviation regulations in design of an

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aircraft.1.2. Statistical data collection, processing and analysis. Selection of aircraft main

relative initial parameters (Characteristics).1.3. Selection and grounds of aircraft configuration, type of its power plant.1.4. Selection of engines and examination of take off run.1.5. Determination optimization of aircraft components and design parameters.1.6. Development of design–structural configuration, aircraft center of gravity.1.7. Standard specification of designed aircraft.

Realization of calculations, models and drawings:

master–geometry of aircraft surface, outline drawing (format А1); Design–structural layout of aircraft (format А1).

2. Impact analysis in changes of aircraft component design parameters under their optimization in aerodynamic and weight characteristics of aircraft

2.1. Determination of designed aircraft drag.2.2. Lift force, induced drag, aircraft polar curve, aircraft lift–drag ratio, aircraft

polar.2.3. Longitudinal moment and location of aerodynamic center of aircraft.2.4. Influence of aircraft design parameters on its aerodynamic and weight

characteristics._________________________________________________________________

3. Integrated designing and computer–aided modeling __________SURFACE MODEL__________ of designed aircraft

3.1. Development of unit master–geometry.3.2. Determination of loads acting on unit.

Realization of calculations, models and drawings:

unit master–geometry;

4. Integrated designing and computer–aided modeling of aircraft systems

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4.1. Hydraulic system designing and modeling.4.2. Maintenance Manual of designed system._________________________________________________________________

Realization of calculations, models and drawings:

system schematic diagram (format А2);Technological Section

5. Development of aircraft unit manufacturing technique

5.1. Development of enlarged production (manufacturing) methods in assembly of units: selection of tools and equipment, specifications for delivery of parts and assembly units, development of production charts for assembly procedure, standardization, assembly cycle schedule.

_________________________________________________________________

Realization of calculations, models and drawings:

Economical section

6. Calculation of economic efficiency characteristics

6.1. Business plan: companys history, aircraft characteristic, product market, marketing, personnel and management, risk analysis and their prevention.

6.2. Project financing: sources of financing, receipts and expenditures – calculation of expenditures for designing and manufacturing, calculation of cost value, price income, calculation of companys minimal internal funds, determination of point of make out, calculation of direct and indirect costs.

6.3. Total transportation cost value and company's revenue.6.4. Income from project.6.5. Influence in change of aircraft and its units design parameters on aircraft

efficiency criteria.

7. Special assignment

Cabin layout and interiors design of the aircraft. Seating arrangement with high comfort level ______________________________________________________

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2. Explanatory note contents (list of questions subjected to development):

in compliance with assignment. Design-explanatory

note with Figures, Tables involved in text – up to 120 pages.

3.  List of Graph materials (with obligatory drawings clearly specified):

graph material and presentation in strict correspondence to the assignment

Information on CD–R or DVD+/–R medium installed in department

computer network prior to defense

4. Date of assignment issue:

5. Date of final project presentation:

Project supervisor (Date, signature)

Assignment accepted to fulfillment

« » 200 (Date, students signature)

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2012

CONTENTSABSTRACT ………………………………………………………………………… 4INTRODUCTION………………………………………………………………… 5AIRCRAFT DESIGN PROCESS………………………………………………… 7

GENERAL DESIGNING OF AIRCRAFTPURPOSE OF THE AIRCRAFT………………………………………………….11REQUIREMENTS FOR FLIGHT PERFORMANCES………………………….14DESIGN CHART OF THE DESIGNED AIRCRAFT……………………………15PROTOTYPE DATAS………………………………………………………………17SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERS…..24CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE AND ITS RESULTS……………………………………………………………………………..25ZERO APPROXIMATION…………………………………………………………31STATISTICAL COMPUTATION OF MASSES OF AIRCRAFT………………..32AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERS…………………33SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATION…………..43SELECTION OF ENGINE…………………………………………………………..55AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICAL METHOD……………………………………………………………………………….63MAXIMUM TAKE-OFF MASS………………………………………………………66CENTER OF GRAVITY………………………………………………………………68DESIGN STRUCTURAL CONFIGURATION …………………………………….83

AERODYNAMICSAIRCRAFT DESIGN PARAMETERS ON AERODYNAMIC CHARACTERISTICS………………………………………..83CALCULATION OF AERODYNAMIC PARAMETERS USING THE SOFTWARE…………………………………………92CALCULATION OF ZERO DRAG COEFFICIENT FOR TAKE-OFF AND LANDING…………………………….109

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INTEGERATED DESIGN OF AIRCRAFT AND LOAD CALCULATIONAIRCRAFT MASTER GEOMETRY USING UNIGRAPHICS………………….114WING LOAD CALCULATION……………………………………………………..118CALCULATION OF THE DISTRIBUTED FUEL LOAD ON A PLANE WING…119THE WING STRUCTURE MASS LOAD ALLOCATION…………………………125SHEAR FORCE, BENDING MOMENT AND REDUCED MOMENT……………127CALCULATION SCHEME OF REDUCED MOMENT FROM CONCENTRATED LOADS AND FROM ALL LOADS……………………………………………………132

AIRCRAFT SYSTEMS DESIGN AND SCHEMATIC LAYOUTAIRCRAFT HYDRAULIC SYSTEM……………………………………….134HYDRAULIC FLUID…………………………………………………………135COMPONENTS INVOLVED IN HYDRAULIC SYSTEM…………………137HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFT……….151HYDRAULIC SYSTEM PANEL……………………………………………..154HYDRAULIC SYSTEM MAINTENANCE………………………………….157

MANUFACTURING TECHNOLOGY OF VERTICAL RIBAIRCRAFT RIB……………………………………………………………………..159RIB CONSTRUCTION……………………………………….……………………..163PRODUCTION METHOD OF PARTS OF THE RIB……………………………166ASSEMBLY PROCEDURE OF THE RIB…………………………………………168STAGES OF FORMATION OF RIB DIMENSIONS USING TEMPLATES……171

AIRCRAFT VERTICAL ASSEMBLY JIG DESIGN LAYOUT DIAGRAM…….173

ECONOMICAL SECTIONECONOMIC EFFICIENCY CHARACTERISTICS CALCULATION………..174

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SPECIAL ASSIGNMENTINTERIOR CABIN LAYOUT AND SEATING ARRANGEMENT……..…………..179FULFILLING REQUIREMENTS OF THREE ABREAST SEATING LAYOUT….179.CABIN DIMENSIONING FOR 3- ABREAST SEATING…………………………..…179INTERIOR ARRANGEMENT – CROS SSECTION (TYPICAL)……………………180DETERMINATION OF DESIGNED AIRCRAFT CABIN CROSS-SECTION…......181DETERMINATION OF CABIN LENGTH FOR HIGH COMFORT LEVEL………183

3-VIEW DRAWING OF THE AIRCRAFT …………………………………189DESIGN STRUCTURAL LAYOUT………………………………………….188CENTRE OF GRAVITY LOCATION OF THE AIRCRAFT……………..188HYDRAULIC SYSTEM SCHEMATIC OF THE AIRCRAFT…………….190CABIN SEATING LAYOUT OF THE AIRCRAFT………………………..189

LIST OF DIAGRAMS

CONCLUSION …………………………………………………………………. 191

REFERENCE ………………………………………………………………………. 192

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ABSTRACTThe General design of the aircraft is carried out on basis of collection of aircraft

statistical data and in accordance with the pilot project development task and finally the general view of an aircraft is presented. The main purpose of the aircraft design requirement is fulfilled according to the aviation rules and regulations.

The main project is categorized into five part, in the first part aircraft take-off mass in zero approximation is determined and followed by designing the weight of main units, fuel, equipment, control system, geometrical dimensions of a wing, tail units, fuselage, landing gear, location of their center of masses, calculating the aircraft’s center-of-gravity. Finally the design specifications for the aircraft are presented.

The second part is focused on determining aerodynamic forces acting upon the designed aircraft.

The third part is the development of aircraft unit structure using Computer aided designing. The loads acting on the designed aircrafts unit structure is calculated and the materials for unit structure are selected. The fourth part is concerned about the systems developed for designed aircraft. The schematic layout of the hydraulic system and its purpose are briefed following the operations & maintenance manual of the designed system. The last part is the technological section where the development of production charts for assembly of designed aircraft vertical rib structure is done. In special assignment the seating arrangement of the designed aircraft is sketched and interiors of aircraft components are briefed with the current industry techniques.

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INTRODUCTIONThe purpose of designing a new aircraft is the creation of a structure with unique characteristics, which should be reliable, economical fulfilling the conditions of operation, performance requirements and its primary goal should be attained. To perform the preliminary design structure of the aircraft it is necessary to be knowledgeable in the field of general arrangement of aircraft and helicopters, design of power units and systems, construction of elements of assembly structures and units of the aircraft, aero hydrodynamics, durability, technologies, material science, and economics. The purpose of design is to develop a project, realization of which, being limited to a certain extent, would ensure the most efficient reaching of the defined goals of the design.In designing a new aircraft the following should be considered, fulfillment of targeted tasks stability and controllability of an aircraft on a specified trajectory control and navigation in various flight conditions life support Performance characteristics Characteristics of technological level of the serial aircraft and its economic efficiency The special equipment Standardization and unification level Requirements to reliability and maintenance system Power plant and its systems Perspective of development of the aircraft and its basic systemsAn aircraft is an element of the aviation complex, which seamlessly unites human and

material resources and carries out certain useful functions. The functional-structural diagram of the aviation complex is shown on Figure. The aviation complex is an element of state transport or defense system. All this defines necessity to use systematic approach to aircraft design.

To implement the process of aircraft design, there was necessity to create specialized development design offices, which include complicated laboratory and manufacturing research. The activities of development design officers are based on work of branch-wise research institutes, which research the prospects of aviation development in various directions, and on experience of aircraft production and operation.

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FOUR STAGES OF DESIGNING1) External designing: At this stage the research of complicated organization-technical

systems including an aircraft or aircraft family as an element is carried out.2) The second stage — the development of a technical proposal: At this stage, the

scheme is selected and optimal combination of basic aircraft parameters, composition and structure of systems ensuring fulfillment of required functions is determined.

3) Third stage – front end engineering: In the process of design arrangement the aircraft center-of-gravity is specified. The calculation of center-of-gravity is followed by making weight reports on the basis of strength and weight calculations of airframe and power unit, lists of equipment, outfit, cargo etc.

4) The fourth stage – working draft: The purpose of this stage is issuing all technical documentation required for production, assembly, mounting of separate units and systems and the whole aircraft as well. At this stage, on the basis of design-technological elaboration the drawings with general view of aircraft units, assembly and working-out drawings of separate parts of the aircraft.

MAIN STAGES OF AIRCRAFT PROJECT DEVELOPMENT

AIRCRAFT DESIGN PROCESSThe aircraft design process is the steps by which aircraft are designed. These depend on many factors such as customer and manufacturer demand, safety protocols, physical and economic constraints etc. For some types of aircraft the design process is regulated by national airworthiness authorities. This article deals with powered aircraft such as airplanes and helicopter designs.Aircraft design is a compromise between many competing factors and constraints and accounts for existing designs and market requirements to produce the best aircraft.

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DESIGN CONSTRAINTS IN DESIGNING PROCESSA. Aircraft regulationsAnother important factor that influences the design of the aircraft are the regulations put forth by national aviation airworthiness authorities.Airworthiness Certificates-An airworthiness certificate is an FAA document which grants authorization to operate an aircraft in flight.Standard Airworthiness Certificate-A standard airworthiness certificate (FAA form 8100-2 displayed in the aircraft) is the FAA's official authorization allowing for the operation of type certificated aircraft in the following categories:

Normal Utility Acrobatic Commuter Transport Manned free balloons Special classes

FUNCTIONAL-STRUCTURAL CHART OF THE AVIATION COMPLEX

A standard airworthiness certificate remains valid as long as the aircraft meets its approved type design, is in a condition for safe operation and maintenance, preventative maintenance, and alterations are performed in accordance with 14 CFR parts 21, 43, and 91.Airworthiness Certification Process-The FAA requires several basic steps to obtain an airworthiness certificate in either the Standard or Special class.The FAA may issue an applicant an airworthiness certificate when:

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o Registered owner or operator/agent registers aircraft, o Applicant submits application (PDF) to the local FAA office, ando FAA determines the aircraft is eligible and in a condition for safe operation

A. Environmental factorsAn increase in the number of aircraft also means greater carbon emissions. Environmental scientists have voiced concern over the main kinds of pollution associated with aircraft, mainly noise and emissions. Aircraft engines have been historically notorious for creating noise pollution and the expansion of airways over already congested and polluted cities have drawn heavy criticism, making it necessary to have environmental policies for aircraft noise. Noise also arises from the airframe, where the airflow directions are changed. Improved noise regulations have forced designers to create quieter engines and airframes. Emissions from aircraft include particulates, carbon dioxide (CO2), Sulphur dioxide (SO2), Carbon monoxide (CO), various oxides of nitrates and unburnt hydrocarbons. To combat the pollution, ICAO set recommendations in 1981 to control aircraft emissions. [11] Newer, environmentally friendly fuels have been developed and the use of recyclable materials in manufacturing have helped reduce the ecological impact due to aircraft. Environmental limitations also affect airfield compatibility. Airports around the world have been built to suit the topography of the particular region. Space limitations, pavement design, runway end safety areas and the unique location of airport are some of the airport factors that influence aircraft design. B. SafetyThe high speeds, fuel tanks, atmospheric conditions at cruise altitudes, natural hazards (thunderstorms, hail and bird strikes) and human error are some of the many hazards that pose a threat to air travel. Airworthiness is the standard by which aircraft are determined fit to fly.[19] The responsibility for airworthiness lies with national aviation regulatory bodies, manufacturers, as well as owners and operators. The International Civil Aviation Organization sets international standards and recommended practices for national authorities to base their regulations on The national regulatory authorities set standards for airworthiness, issue certificates to manufacturers and operators and the standards of personnel training. Every country has its own regulatory body such as the Federal Aviation Authority in USA, DGCA (Directorate General of Civil Aviation) in India, etc.

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C. Design optimizationAircraft designers normally rough-out the initial design with consideration of all the constraints on their design. Historically design teams used to be small, usually headed by a Chief Designer who knew all the design requirements and objectives and coordinated the team accordingly. As time progressed, the complexity of military and airline aircraft also grew.D. Design aspectsThe main aspects of aircraft design are:

1. Aerodynamics2. Propulsion3. Controls4. Mass5. Structure

All aircraft designs involve compromises of these factors to achieve the design mission.

E. Computer-aided design of aircraftIn the early years of aircraft design, designers generally used analytical theory to do the various engineering calculations that go into the design process along with a lot of experimentation. These calculations were labor intensive and time consuming. In the 1940s, several engineers started looking for ways to automate and simplify the calculation process and many relations and semi-empirical formulas were developed. Even after simplification, the calculations continued to be extensive. With the invention of the computer, engineers realized that a majority of the calculations could be done by computers, but the lack of design visualization and the huge amount of experimentation involved kept the field of aircraft design relatively stagnant in its progress.F. Financial factors and marketBudget limitations, market requirements and competition set constraints on the design process and comprise the non-technical influences on aircraft design along with environmental factors. Competition leads to companies striving for better efficiency in the design without compromising performance and incorporating new techniques and technology.

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DESIGN SECTION PART-1COMPUTER AIDED GENERAL DESIGNING OF AIRCRAFT

1. PURPOSE OF THE AIRCRAFT2. REQUIREMENTS FOR FLIGHT PERFORMANCES3. DESIGN CHART OF THE DESIGNED AIRCRAFT4. PROTOTYPE DATAS5. SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERS6. CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE

AND ITS RESULTS7. ZERO APPROXIMATION8. STATISTICAL COMPUTATION OF MASSES OF AIRCRAFT9. AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERS10. SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATION11. SELECTION OF ENGINE12. DETERMINATION OF CENTER OF GRAVITY OF THE AIRCRAFT

13. AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICAL METHOD

14. MAXIMUM TAKE-OFF MASS15. CENTER OF GRAVITY16. DESIGN STRUCTURAL CONFIGURATION17. DESIGN SPECIFICATION

PURPOSE OF THE DESIGNING AIRCRAFTA. Aircrafts Intended Purpose - Commercial usage

Commercial usage denotes using the aircraft for a business purpose or getting

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directly/indirectly financial gain from it.

B. Payload category - PassengersAircraft adapted for carrying passengers.

C. Type - Regional jet

The term regional jet describes a range of short to medium-haul turbofan powered aircraft, whose use throughout the world expanded after the advent of airline deregulation in the United States in 1978.

ExamplePRIMARY USERS MANUFACTURER ROLEAeroflot Yakolev Yak-40 regional sized mini-jet

airlinersSkyWest AirlinesPinnacle AirlinesExpressJetComair

Bombardier CRJ100 Regional jet/Business jet

Aerosvit AirlinesRossiya

Antonov An-148 regional jet

D. Range - Short-range short range refers to distance travelled is between 2500.2 km (less than 1350nm) and Time taken to travel is less than 5 hours

ExampleFROM & TO DISTANCE in km TIMENew York-Miami 2051.914 2 hours 49 minsTokyo-Seoul 1,159.04 1.5 to 2 hoursDenver-Boston 2800 3hrs 42 mins

G. Special Requirements - Cargo Carrying capabilityCan be used to carry cargos and can be used as a cargo variant

H. Mode of Class - Economy class Economy class refers to the seating arrangement of the aircraft which is usually reclined and include a fold-down table. The seats pitch range from 29 to 36 inches (74 to 91 cm), usually 30–32 in (76–81 cm), and 30 to 36 in (76 to 91 cm) for international economy class seats. Domestic economy classes range from 17 to 18.25 in (43 to 46.4 cm).

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GENERAL REQUIREMENTS1. The aircraft, its engines, equipment and other parts, and operational publications shall

meet the following requirements: aviation requirements АП-25 and additional requirements for airworthiness of

"AIRCRAFT NAME" aircraft, in consideration of its design and operational features, forming the "Certification basis of aircraft of "AIRCRAFT NAME" type" together with mentioned requirements;

engine - aviation requirements АП-33; APU - aviation requirements АП-ВД.

2. As for engine emission the aircraft shall meet the requirements of Appendix 16 to International Aviation Convention (Volume II «Aviation engines emission», Edition 1981, Revisions 1 to 4) and requirements of Aviation Regulations АП34.

3. As for protection against hijacking the aircraft shall meet the requirements ICAO Appendix 6,8,17 (with Revisions 97 and 98)Ukrainian Air Law (Section 8).

4. Processing and analysis of flight data using the ground personal computer shall be provided to control the correctness of maintaining of preset flight modes and the pilot technique, to evaluate the pilots' professional level, technical state of the aircraft, its equipment and functional systems in monitoring of operation conditions within life time limits.

The system shall include:

aircraft removable data carrier, receiving the information from corresponding aircraft signal transmitters;

personal computer with printer, input and reproducing device and specific software.

5.Ground facilities and repair equipment shall correspond to this performance specification.

6.Simulators and training devices should be designed for aircraft according to individual performance specifications. The programs for training of flight and technical staff should be developed up to completion of certification tests.

SPECIFIC AIRCRAFT STRUCTURE REQUIREMENTS

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. The airplane should be designed and manufactured by a principle of “fail-safe structure”.

Weight layout and airplane center-of-gravity should ensure a capability of operational both with total and short number of passengers at all possible operational versions of loading and fuelling according to the instruction of loading and centre-of-gravity not using ballast. Limit of on-ground tail-heavy center of gravity be no less than 5 % of MAC.

The capability of creation of convertible and transport versions should be provided on the basis of this airplane according to special performance specification.

REQUIREMENTS FOR FLIGHT PERFORMANCESMaximum passenger capacity with distance between the seats 750 (762) mm,

person 55

Maximum payload

kg 5000Cruise speed:at long range cruisemaximum

km/h 835

Cruise altitude,

km 10.5Required length of RWY (SA, Н = 0, dry concrete), for takeoff:for landing:

m 19502250

 Applied flight range (emergency fuel reserve for 0.75 hour of flight; takeoff in SA conditions; Н = 0)with maximum payload

km 2500

Fuel consumption for 1 pass/km while flying for technical range with maximum payload

g 340

Maintenance and overhaul, manhour

8.8

REQUIREMENTS FOR ENVIRONMENTAL PROTECTION

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. As for perceived noise the aircraft should meet the requirements of Chapter 4 of "Environmental protection" International Standards, Appendix 16 to the International Civil Aviation Convention (Volume I «Air noise», 2001) and to requirements of Part 36 of Aviation Regulations АП-36.

To decrease atmospheric pollution and reduce fuel flow at ground operation the capability of fulfillment of taxiing before take-off and after landing with one operating engine should be worked out on airplane.

DESIGN CHART OF THE DESIGNED AIRCRAFT

STATISTICAL DATA COLLECTION Statistical data collection is the process of collecting flight, mass, power plant and geometrical data’s of required prototypes for the design project. In this project I have

General Design

Collection and process of statistical dataDesign specification and three view diagram

Aerodynamic Characteristics Designed aircraft drag

Design structural unit Calculation of loads acting on unitModelling of designed unit

Systems Design Schematic layout of the hydraulic system

Technological Activity Design of assembly jigs for developed unit

Special activity Cabin layout

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collected four different aircraft data’s and their features are explained and tabulated. These aircrafts are selected based upon the design requirements and design specification mentioned below,

TACTICAL TECHNICAL REQUIREMENTS OF THE DESIGNING AIRCRAFT

Maximum speed , Vmax 835 km/hCruising speed, Vcruise 760 km/hCruising height, Нcruise 11 kmNumber of passengers, npass 47Number of crew members, ncrew 4Range, L 2000 kmTake-off distance, Lр 1970 mVertical speed, Vy 14 m/sMaximum take-off weight, Ммах 40 tons

DESIGN SPECIFICATION OF THE AIRCRAFT

Type of the aircraft - Transport category with capacity to carry 47 to 55 passengers including crew

Aerodynamic configuration Normal configuration with horizontal stabilizer on tail section

Wing Low wing with Dihedral and wing sweepTail T-tail configurationFuselage Cylindrical shapePower plant type Turbofan located at aft part of the fuselageLanding gear Tricycle configuration with nose wheel

Based on the tactical technical requirements and the general design specification of the designing aircraft we are gathering the similar aircrafts and their detailed specification is tabulated. From the critical parameters of the aircraft are listed. With the obtained results we now ready to input all parameters in the software which would give all the relative masses and some important parameters for further calculation.Aircrafts data are gathered from various sources which include books, magazines, websites, etc., Some of the missing parameters are found manually by calculations or it can be found by scaling the three view picture of the collected aircraft. In obtaining the details it is important to have the three view pictures of each aircraft for simplification further in drawing the designed aircraft three view it is very helpful .A short brief of the aircraft is provided for each of the aircraft with its variant and their three view picture.

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Upon the four aircrafts selected we can take any one from that as a main prototype for further simplification. I have selected the following aircrafts for comparison, 1.EMBRAER ERJ 145 2. BOMBARDIER CRJ100 3. TUPOLEV 134-A 4. BOEING 717-200

My main prototype is EMBRAER ERJ 145

AIRCRAFTS SELECTED FOR STATISTICAL DATA COLLECTION AND THEIR PARAMETERS

EMBRAER ERJ 145

The Embraer ERJ 145 family is a series of regional jets produced by Embraer, a Brazilian aerospace company. Family members include the ERJ 135 (37 passengers), ERJ 1 (44passengers), and ERJ 145 (50 passengers). The key features of the production design included:

1. Rear fuselage-mounted engines2. Swept wings (no winglets)3. "T"-tail configuration4. Range of 2500 km

Civilian models ERJ 135ER - Extended range, although this is the Baseline 135 model. Simple shrink

of the ERJ 145, seating thirteen fewer passengers, for a total of 37 passengers. ERJ 135LR - Long Range - increased fuel capacity and upgraded engines.

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ERJ 140ER - Simple shrink of the ERJ 145, seating six fewer passengers, for a total of 44 passengers.

ERJ 140LR - Long Range (increased fuel capacity (5187 kg) and upgraded engines. ERJ 145STD - The baseline original, seating for a total of 50 passenger

Military models C-99A - Transport model EMB 145SA (R-99A) - Airborne Early Warning model EMB 145RS (R-99B) - Remote sensing model

BOMBARDIER CRJ100

The Bombardier CRJ100 and CRJ200 are a family of regional airliners manufactured by Bombardier, and based on the Canadair Challenger business jet.

The CRJ100 was stretched 5.92 meters (19 feet 5 inches), with fuselage plugs fore and aft of the wing, two more emergency exit doors, plus a reinforced and modified wing. Typical seating was 50 passengers, the maximum load being 52 passengers. The CRJ100 featured a Collins ProLine 4 avionics suite, Collins weather radar, GE CF34-3A1 turbofans with 41.0 kN (4,180 kgp / 9,220 lbf), new wings with extended span, more fuel capacity, and improved landing gear to handle the higher weights. It was followed by the CRJ100 ER subvariant with 20% more range, and the CRJ100 LR subvariant with 40% more range than the standard CRJ100. The CRJ 100 SE sub-variant was produced to more closely meet the needs of corporate and executive operators.Variants

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Several models of the CRJ have been produced, ranging in capacity from 40 to 50 passengers. The Regional Jet designations are marketing names and the official designation is CL-600-2B19.CRJ100 -The CRJ100 is the original 50-seat version. It is equipped with General Electric CF34-3A1 engines. Operators include Jazz Aviation, Comair and more.

CRJ200 -The CRJ200 is identical to the CRJ100 except for its engines, which were upgraded to the CF34-3B1 model, offering improved efficiency.

CRJ440 -Certified up to 44-seat, this version was designed with fewer seats in order to meet the needs of some major United States airlines.Challenger 800/850 - A business jet variant of the CRJ200

TUPOLEV 134-A

The Tupolev Tu-134 (NATO reporting name: Crusty) is a twin-engined airliner, similar to the French Sud Aviation Caravelle and the later-designed American Douglas DC-9, and built in the Soviet Union from 1966–1984. The original version featured a glazed-nose design and, like certain other Russian airliners (including its sister model the Tu-154), it can operate from unpaved airfields.

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Design and development

Following the introduction of engines mounted on pylons on the rear fuselage by the French Sud Aviation Caravelle, airliner manufacturers around the world rushed to adopt the new layout. Its advantages included clean wing airflow without disruption by nacelles or pylons and decreased cabin noise. At the same time, placing heavy engines that far back created challenges with the location of the center of gravity in relation to the center of lift, which was at the wings. To make room for the engines, the tailplanes had to be relocated to the tail fin, which had to be stronger and therefore heavier, further compounding the tail-heavy arrangement.

VariantsTu-134 The glass nosed version. The first series could seat up to 64 passengers, and

this was later increased to 72 passengers. The original designation was Tu-124A.

Tu-134A Second series, with upgraded engines, improved avionics, seating up to 84 passengers. All Tu-134A variants have been built with the distinct glass nose and chin radar dome, but some were modified to the B standard with the radar moved to the nose radome.

Tu-134B Second series, 80 seats, radar moved to the nose radome, eliminating the glazed nose. Some Tu-134B models have long-range fuel tanks fitted under the fuselage; these are visible as a sizeable bulge.

Tu-134UBL

Bomber aircrew training version.

Tu134UBK Naval version of Tu-134UBL. Only one was ever built.

BOEING 717-200Boeing 717 was specifically designed for the short-haul, high frequency 100-passenger airline market. The highly efficient 717 concluded its production run in May 2006, though the airplane will remain in service for years to come.Final assembly of the 717 took place at the Boeing plant in Long Beach, Calif. The airplane was originally part of the McDonnell Douglas airplane family and designated the MD-95 prior to merger with The Boeing Co. in 1997. The program produced 156 717s and pioneered breakthrough business and manufacturing process for Boeing Commercial Airplanes

[24]

The.The standard 717 has a two-class configuration with 106 seats. Its passenger-pleasing interior features a five-across-seating arrangement in economy class, with illuminated handrails and large overhead stow bins.The two-crew flight deck incorporates six interchangeable liquid-crystal-display units and advanced Honeywell VIA 2000 computers.Flight deck features include an Electronic Instrument System, a dual Flight Management System, a Central Fault Display System, and Global Positioning System. Category IIIb automatic landing capability for bad-weather operations and Future Air Navigation Systems are available.Two advanced Rolls-Royce 715 high-bypass-ratio engines power the 717. The engine is rated at 18,500 to 21,000 pounds of takeoff thrust, with lower fuel consumption and significantly lower noise and emission levels than the power plants on comparable airplanes.DESIGNThe 717 features a two-crew cockpit that incorporates six interchangeable liquid-crystal-display units and advanced Honeywell VIA 2000 computers. The cockpit design is called Advanced Common Flight deck (ACF) and is shared with the MD-11.

STATISTICAL TABLEFLIGHT DATA:

[25]

Flight includes Vmax – the maximum speed of flight; HV max – flight altitude with the maximum speed; Vcruise –cruise speed; Нcruise –cruise altitude; Vland – landing speed; Vto – take-off speed; VY– rate of climb; Hclg– static ceiling; L– flight range; Ltor – distance of the take-off run; Lto – take-off distance; Lroll–landing roll distance; Lland– landing distance;

FL

IGH

T D

AT

A

1 No 1 2 4 32 Name of the

aircraftProducerCountryYear of production

EMBRAER 145EmbraerBrazil1989-present

Bombardier CRJ200 BombardierCanada1992

TUPOLEV 134-A

TupolevSoviet union1966–1984

BOEING 717-200BoeingUnited stated1998–2006

3 Source Janes all the world aircraft and Wikipedia4 Vcruise, km/h 833 850 850 8115 Vmax, km/h 679 785 950  6296 Нcruise, km 11.277 11 11 10.4007 HV max, km 9.753 11 11.5 11.2808 Vto, km/h 170 155 - 1509 Vland, km/h 233 250 - 24410 VY, km/h 6.5 6 - 611 Hclg, km 11.27 12.49 12.1 1112 L(mf Max ) , km 3037 2500 - 264513 L(mcargo max) ,

km2963 1800 1020 3800

14 Lto, km 1.97 1527 2.4 1.715 Lland, km 1.3 1423 2.2 1.52

MASS DATA : This includes take-off mass(m0), maximum take-off mass(m0max), payload mass(mpld), number of passengers(npass), landing mass(mland), empty mass(mempty), mass of crew(mc), mass of fuel(mf), empty equipped mass(mempt.eqpd) and total mass(mtotal ).

[26]

MA

SS

DA

TA

S.NO MASS EMBRAER 145

Bombardier CRJ200

TUPOLEV 134-A

BOEING 717-200

16 m0 (mto), kg 19200 21636 47000 4989517 m0max, kg 20000 22000 47200 2200018 mpld, kg 5640 6240 8200 1200019 npass 47 52 84 10020 mland, kg 18700 20000 43000 4335921 mempty, kg 11585 19,958  27,960 3000022 mc, kg 5,284  6,124  8,200 1240023 mf, kg 2865 4300 - 850024 mempt.eqpd, kg 17,100  13730 29050 4354525 mtotal, kg 20,100 24,041 47,600 49,900 

POWERPLANT DATA: This includes engine thrust (P0), mass of engine (meng), number of engines and its type, specific fuel consumption (Cp) and bypass ratio(Y).

PO

WE

RP

LA

NT

DA

TA

S.NO

ENGINESPECS

EMBRAER 145

Bombardier CRJ200

TUPOLEV 134-A

BOEING 717-200

26 P0 (N0), daN (kN)

31.3 31 103 97.9

27 meng, kg 1438 2305 464028 No of

enginesType of engine

2 2 2Twin-spool non-afterburning turbofan

2

29 Cp, lb/lbf·hr 0.39 - 0.498 -30 Y, Bypass

ratio3:1 5:1 - -

GEOMETRICAL DATA: This includes wing area(S), wing span(L), sweep angle(),

aspect ratio of wing(), thickness ratio at chord(c̄0 ) and at tip(c̄ tip ),taper ratio(), length of

[27]

fuselage(Lf), diameter of fuselage(df), area of aileron(Sail ), relative fuselage mid section

area(ΣSmcs ), wing loading(P0) and thrust to weight ratio(t0).

GE

OM

ET

RIC

AL

DA

TA

S.NO GEOMETRICALPARAMETERS

EMBRAER 145

Bombardier CRJ200

TUPOLEV 134-A

BOEING 717-200

31 S, m2 51.18 54.54 127.3 92.9732 L, m 20.04 20.52 29.00 28.4533 22.73 24.75 35.00 24.5034 7.85 7.72 6.61 8.735 c̄0

c̄ tip

4.091.04

5.131.27

--

--

36 4 3.4 0.255 5.1037 Lf, m 29.87 24.38 37.10 3338 df, m 2.28 2.69 2.9 3.3439 f 12.25 9.06 11.45 4.3040 Sail

, m21.70 1.93 - -

41 ΣSmcs , m2 7.56 8.38 10.5416 18.84

42 P0=m0g/10S, daN/m2

375.15 394.3 369.21 556

43 t0=10P0/m0g 0.3326 0.3884 0.289 0.3806

DERIVATIVE VALUE: This includes specific fuel weight (eng), effective load factor (

Keff . load ), relative aileron area ( S̄ail ), relative horizontal ( S̄HT ) and vertical stabilizer area (

S̄VT ).

DE

RIV

AT

IVE

S.NO DERIVATIVEPARAMETERS

EMBRAER 145

Bombardier CRJ200

TUPOLEV 134-A

BOEING 717-200

44 eng, kg/daN2 306.51 263 - 29445

Keff . load=mc arg o

m0

0.2752 0.288 0.1744 0.2405

46 Kmcs=m0/ΣSmcs , 2539 2600 4459 2627

[28]

VALUES

daN/m2

47 S̄ail=Sail /S 0.0332 0.0353 - -

48 S̄HT=SHT /S 0.219 0.173 0.241 0.205

49 S̄VT=SVT / S 0.141 0.168 0.167 0.210

SELECTION OF AIRCRAFT MAIN RELATIVE INITIAL PARAMETERSThus finally tabulating all the required values it is necessary to find the main relative initial parameters of wing, fuselage and the tail unit. The obtained result is used in the software. WING PARAMETERaspect ratio, 7.85sweep angle, 22.73taper ratio, 4

relative width of airfoil, c̄ 18 or 0.18

relative chord of flap, b̄ f=b f /bwing

0.25

deflection angles of flap, f 18

relative area of ailerons,S̄ail=Sail /S 0.06

FUSELAGE PARAMETERfineness ratio f 12.25fuselage diameter Df 2.50

TAIL UNIT PARAMETER

relative area of horizontal stabilizer, S̄HT=SHT /S 0.219

relative area of vertical stabilizer, S̄VT=SVT / S 0.141

aspect ratio of horizontal surface, HS 4.077aspect ratio of vertical surface, VS 1.36Sweep angle of horizontal surface, HS 20Sweep angle of horizontal surface, HS 32

Relative thickness of horizontal surface, c̄ HS12 or 0 .12

Relative thickness of vertical surface, c̄ VS10 or 0.10

[29]

CALCULATION OF AIRCRAFT MASSES THROUGH THE SOFTWARE AND ITS RESULTS

Aircraft masses in zero approximation are calculated using software by entering necessary parameters taken from statistical data and the initial parameters. First the relative masses for fuselage, wing, power plant, tail unit, fuel and landing gear are found with respect to the aspect ratio of wing taken as 4 and aircrafts wing loading attained from graphical result as 600 N/m2. The graphs obtained from this result help us to select the desired wing loading and from the wing the lowest value of takeoff mass is taken as the final one. The relative masses are changed to direct masses by multiplying it with the finally obtained take-off mass, for me it is 39.11tons. Therefore my relative masses are multiplied with 39.11 ton to get direct mass

Graphs for each lab are plotted versus each parameter and from that the final wing loading is obtained followed by the takeoff mass. Other parameters obtained include engine performance data like thrust to weight ratio at take-off, climbing and cruise. My main comparative parameter is Aspect ratio which I took in three variations as

ASPECT RATIO 2 , 4 AND 6

AIM OF THE LAB AND ITS RESULT TO FIND THE RELATIVE MASS OF AIRCRAFT

LAB 5: In this part we are finding the relative mass of power plant respect to the aspect ratio of wing taken as 4 and aircrafts wing loading 600 N/m2.

[30]

RESULT: In table P,denotes wing loading Tk,aspect ratio and SU is the RELATIVE MASS OF POWER PLANT which is 0.086

LAB 7A:In this part we are finding the relative mass of wing respect to the aspect ratio of wing taken as 4 and aircrafts wing loading 600 N/m2.

RESULT: In table p, denotes wing loading and Tk, aspect ratio and Mkp is the RELATIVE MASS OF wing which is 0.048.

[31]

LAB 7B:In this part we are finding the relative mass of fuselage respect to the aspect ratio of wing taken as 4 and aircrafts wing loading 600 N/m2.

RESULT: In table DF, refers to diameter of the fuselage (3.84m) and Lf refers to aspect ratio of the fuselage (12.25). By comparing both the values we get the relative value of fuselage equals to 0.350.LAB 7G:In this part we are finding the relative mass of tail unit with respect to the aspect ratio of wing taken as 4 and aircrafts wing loading 600 N/m2.

RESULT: In table P, denotes wing loading and MOP, denotes RELATIVE MASS OF THE TAIL UNIT which is 0.0182

[32]

LAB 7V:In this part we are finding the relative mass of landing gear with respect to the aspect ratio of wing taken as 4 and aircrafts wing loading p,600 N/m2.

RESULT: RELATIVE MASS OF THE LANGING GEAR is 0.062

LAB 8: In this part we are finding the mass of equipment, crew and payload.

RESULT: MASS OF THE Equipment, crew and payload is 9502.98kg

LAB 9: In this part we are finding the take off mass of the aircraft which is equal to 39.11tons.

[33]

RESULT: Take off MASS OF is 39110 kg or 39.11 tons

RESULTS FROM GRAPH WITH RESPECT TO THE RELATIVE PARAMETER

LAB NO

PARAMETERS RESULTS

3 Lift to drag ratio 12.204 Thrust to weight ratio at Take-off 0.2514 Thrust to weight ratio at Landing 0.2714 Thrust to weight ratio at Cruising 0.1795 Mean Thrust to weight 0.2715 Relative mass of powerplant 0.0867a Relative mass of wing 0.0487b Relative mass of fuselage 0.3507g Relative mass of Tail unit 0.01827B Relative mass of Landing gear 0.0626 Relative mass of Fuel 0.2248 MEQ = MCREW+MPAYLOAD+MEQUIPMENT 9502.98 kg9 Take-off mass relative to wing loading 39110 kg

DIRECT MASS

[34]

Mass of fuselage 13688.5 kgMass of wings 1877.28 kgMass of tail unit 711.802 kgMass of powerplant 3363.46 kgMass of landing gear 2424.82 kgMass of fuel 760.64 kg

COMPUTATION OF AIRPLANE TAKE-OFF MASS IN ZERO APPROXIMATIONDETERMINATION OF MASS FROM LAB RESULTS Mass of fuselage = Relative mass of fuselage * Take off mass = 0.350 * 39110 kg = 13688.5 kgMass of wings = Relative mass of wing * Take off mass = 0.048* 39110 kg = 1877.28 kgMass of tail unit = Relative mass of tail unit * Take off mass = 0.0182* 39110 kg = 711.802 kgMass of power plant = Relative mass of power plant * Take off mass = 0.086* 39110 kg = 3363.46 kgMass of landing gear = Relative mass of landing gear * Take off mass = 0.062* 39110 kg = 2424.82 kgMass of Fuel = Relative mass of fuel * Take off mass = 0.224* 39110 kg = 8760.64 kgMass of crew = 4 * 80 kg = 320kg Mass of payload = 47 * 90 kg = 4230kg Mass of Equipment and control systems = 4952kg

ZERO APPROXIMATIONTake-off mass of the airplane for zero approximation is determined by the formula received from the equation of mass ratio with statistical data.m0=mst+mp . p+mf+mpl+mcrew+meq ;

[35]

Here,m0 = Take-off mass, mst = Structural mass of the aircraft, mp . p = Power plant mass,

mf = Fuel mass, mpl = Payload mass, mcrew = Crew mass, meq = Equipment mass

Mass Ratio (dimensionless) equation is, 1=mst+mp. p+mf +meq+

mpl+mcrew

m0

Re-arranging we get final takeoff mass as,

m0=

mpl+mcrew

1−(m st+mp. p+mf +meq )

mst - Relative airframe mass = Relative mass of fuselage+ Relative mass of wing+ Relative mass of tail unit+ landing gear = 0.350+0.048+0.0182+0.062 = 0.4782

mp . p - Relative mass of power plant = 0.086

mf - Relative mass of fuel = 0.224

meq - Relative mass of Equipment = 0.1266

m0=4230+3201−(0 . 4782+0 . 086+0 .224+0. 1244 )

m0 = 53403.755 kgSTATISTICAL COMPUTATION OF MASSES OF AIRCRAFT

When the airplane takeoff mass in zero approximation is determined it is necessary to

calculate airframe mass mairfr and its components (mass of the wing

mwing , fuselage mfus ,

tail unit mtail unit , landing gears), and also mass of fuel

mfuel , power plant mpow . pl and

engines meng . Relative masses of airframe, power plant, equipment and control system, and

also of the aircraft performing normal take-off and landing are given in Table below

Plane Purpose m̄ airfr m̄ pow . pl m̄ ctl . sys m̄ fuel

Subsonic passenger long-

distance

light 0.30…0.32 0.12…0.14 0.12…0.14 0,18…0,22

medium 0.28…0.30 0.10…0.12 0.10…0.14 0,26…0,30

heavy 0.25…0.27 0.08…0.10 0.09…0.11 0,35…0,40

Multipurpose for local airlines 0.29…0.31 0.14…0.16 0.12…0.14 0.12…0.18

[36]

Take off mass from lab 9 = 39110 kgRelative mass of Airframe = 0.30Mass of Airframe = Relative mass of airframe * Take off mass = 0.30 * 39110 kg = 12515.2 kgRelative mass of power plant = 0.14Mass of power plant = Relative mass of power plant * Take off mass = 0.14* 39110 kg = 5475.4 kgRelative mass of control systems and equipments = 0.11209Mass of control sys = Relative mass of control systems and equipments * Take off mass = 0.11209* 39110 kg = 4383.839 kgRelative mass of fuel = 0.18Mass of fuel = Relative mass of fuel * Take off mass = 0.18* 39110 kg = 7039.8 kg

AIRCRAFT OPTIMIZATION AND DESIGN PARAMETERSGeometrical parameters for designed aircraft are calculated by formulas taken from pilot project book and rest is determined statistically by comparing with the prototypes. After calculating the geometrical parameters we are drawing the theoretical drawing. The geometrical parameters are calculated and obtained satisfying the general requirements of the aircraft. The stages of aircraft optimization include the following:

Determination of Wing Parameters Determination of Fuselage Parameters Determination of Tail Unit Parameters Determination of Position of Center of Mass of the Airplane Determination of Landing gear parameters

Determination of Wing Parameters:In determining Wing parameters its plan form shape is very important in obtaining number of useful relations that apply to a trapezoidal shape. These are based on knowing the wing area, aspect ratio, taper ratio, and leading-edge sweep angle.Before finding the wing area it is necessary to determine the wing loading corresponding to take off mass of 39110kg, which is found in LAB 9 as given below,

[37]

RESULT: The wing loading is found to be 600 daN/m2

Wing statistical parameteraspect ratio, sweep angle, taper ratio,

7.85 22.73 4

WING AREA

S=m0⋅g10⋅p0 Wherem0=39110 (kg )

, g=9 .8 (m /s2)

,p0=600( dN /m2 )

S=39110×9 .810×600

=63 . 87(m2 )

Wing Span (l ) Where λ= 7.85 (choosing from table)

= 22.39m

Wing Chords (b)

broot=b0=SL⋅( 2 η

η+1 ) Where = 4

[38]

broot=b0=63 .8722. 39

⋅( 2×44+1 )=4 . 5(m)

b tip=b0

η=4 . 5

4=1 .1(m )

Quarter chord line sweep angle of the Wing ( χ0 . 25)χ0 .25=22 .730

(Choosing from the statistic’s table)

Leading edge sweep angle of the Wing ( χ0 )

tg χ0=tg χ 0.25+η−1

λ⋅(η+1 )=tg 22.730+ 4−1

7 .85×(4+1 )=0 .4953

χ0=arctg (0 . 4953 )=26 .3490

Mean Aerodynamic Chord of the Wing (MAC =b Aw)

b A=23⋅b0⋅

η2+η+1η⋅(η+1 )

b A=23×4 .5×42+4+1

4⋅(4+1 )=3. 15(m )

Vertical distance between horizontal central line to MAC (z A )

z A=L6⋅η+2

η+1=22. 39

6×4+2

4+1=4 . 478(m)

Horizontal distance between wing root tips to MAC root

x A=l6⋅η+2

η+1⋅tg χ 0=

22.396

.4+24+1

0 . 4953=2 .2179(m)

Determination of Fuselage ParametersThe size and shape of subsonic commercial aircraft are generally determined by the

number of passengers, seating arrangements and cargo requirements. Seating arrangements on commercial passenger aircraft vary depending on the size and range.

Fuselage fineness ratio f fuselage diameter Df Nose section

fineness ratio , λN

Rear section

fineness ratio, λT

12.25 2.5 m 1.5 2.5

Overall Fuselage length, Lf

[39]

Lf= λf×Df=12 .25×2 .5=30 .62(m)

Fuselage nose length, Lf . n

Lf . n=λN×Df=1 .5×2 .5=3 .75(m)

Fuselage rear length, Lf . r

Lf . r=λT×D f=2 .5×2. 5=6 .25(m)

Fuselage middle section length, Lf . m

Lfm = Lf – Lf.n – Lf.r = 30.62 - 3.75 - 6.25 = 20.62 m

Determination of Tail Unit Parameters

Tail unit parameters include Horizontal and Vertical stabilizer. Their geometrical parameters are determined by the same formulae which were used for the wing.

Horizontal stabilizer parameter

Horizontal stabilizer statistical parameteraspect ratio,

sweep angle, taper ratio, Relative horizontal

stabilizer area, Sh . t

4.077 20 2 0.219

Horizontal stabilizer area Sh . t ==Sh .t×S=0 . 219×63 . 87(m2 )=13 . 98 m2

Where S =63.89, wing area

Length of the Horizontal stabilizer

Lh .t=√λh. t⋅Sh. t=√4 .077×13 .98=7 .55(m)

Chords of the Horizontal tail Unit

broot=b0. ht=Sht

Lht

⋅ 2⋅ηη+1

=13 . 987 .55

×2×2 .2+1

=2 . 48(m)

b tip.ht=b0.ht

ηht

=2.482 .0

=1.24 (m)

[40]

Quarter chord line sweep angle of the Horizontal tail unit ( χ0 . 25 . ht )χ0 .25 .ht=200

(Choosing from the static’s table 1.1)

Mean Aerodynamic Cord of the Horizontal Tail (b A . ht )

b A . ht=23⋅b0 . ht⋅

η2ht+ηht+1

ηht⋅(ηht+1 )=

23×2. 48×

2.2+2.+12 (2+1 )

=1 . 93(m)

Vertical Distance between horizontal central line to MAC (z A . ht )

z A .ht=Lht

6⋅

ηht+2

ηht+1=7 .55

6×2+2

2+1=1 .68 (m)

Leading edge sweep angle of the Horizontal stabilizer ( χ0 )

tg χ0=tg χ 0.25+η−1

λ⋅(η+1 )=tg 200+ 2−1

4 .077×(2+1 )=0. 44572

χ0=arctg (0 . 44572 )=24 . 023 0

Horizontal distance between wing root tip to MAC root (x A . h. t )

x A . ht=Lht

6⋅

ηht+2

ηht+1⋅tg χ 0=

7 . 556

.2+22+1

.0 .4452=0 .7488(m)

Vertical stabilizer parameterVertical stabilizer statistical parameter

aspect ratio,

sweep angle, taper ratio, vs Relative horizontal

stabilizer area, Sh . t

1.36 32 1 0.141

x A .ht=0.30(m )

[41]

Vertical stabilizer area

Sv . t == Sv .t×S=0 . 141×63 . 87(m2 )=9 .01 m2 Where

S =63.89, wing area

Length of the Vertical stabilizer

Lv .t=√ λv .t⋅Sv . t=√1. 36×9 . 01=3 .50 (m) Chords of the Vertical tail Unit

broot=b0. vt=Svt

Lvt

⋅ 2⋅ηη+1

=9 . 013 .5

×1×2.1+1

=2 .57 (m)

b tip.vt=b0 .vt

ηvt

=2 .571=2.57(m)

Quarter chord line sweep angle of the Vertical tail unit ( χ0 . 25 . vt )

χ0 .25 .vt=320 (Choosing from the static’s table 1.1)

Mean Aerodynamic Cord of the Vertical Tail (b A . vt )

b A . vt=23⋅b0 . vt⋅

η2

vt+ηvt+1

ηvt⋅(ηvt+1 )=

23×2 . 57×

12+1+11 (1+1 )

=2 .57 (m)

Vertical Distance between horizontal central line to MAC ( y A . v . t )

y A . v . t=Lv . t

3⋅

ηv . t+2

ηv . t+1=3 . 50

3×1+2

1+1=1 .76 (m)

Leading edge sweep angle of the Vertical stabilizer ( χ0 )

tg χ0=tg χ 0.25+η−1

λ⋅(η+1 )=tg 320+ 1−1

1 .36×(1+1 )=0 .62

χ0=arctg (0 .62 )=31. 790

Horizontal distance between wing root tip to MAC root (x A . v .t )

x A . v .t=Lvs⋅tg χ0=3 .50×tg χ 0=3. 50×0. 62=2. 17(m)DETERMINATION OF LANDING GEAR PARAMETERS

For nose-wheel tri-cycle landing gear the following parameters are considered

Wheel base (b ) : distance between axels of nose wheel and main landing gear wheels in side view. It depends on fuselage length.b  (0.30.5)l fus,

b = 0.48 * 30.62 = 14.6976m

[42]

Nose wheel offset (a ) : It is distance between vertical line passing through the airplane center of gravity and nose wheel axis (or axis of several wheels whenever);a = 0.968 * 14.697 = 14.235m

Main Landing Gear offset (e) : It is distance (on side view) between vertical line passing through the airplane center of gravity and axis (or center line of several wheels, bogie) of MLG;e = 0.031 * 14.697 = 0.462m

Static ground angle() : It is the angle between fuselage construction plane and runway surface. It is generally between 2 to +2. So it is taken as 1.

Angle of wing setting sett : It is the angle formed between wing construction plane to the fuselage axis. It is generally between sett  (04). So it is taken be 2.

Angle of overturning(): It is the angle appearing when fuselage tail part or its tail bump touches the runway surface;Ψ = 2 parking angleamax = 12 maximum angle of attackaw = -1 angle between a wing chord and longitudinal axis of fuselage.Ф = amax –(-1) – 2 = 12+1-2 = 11As a rule  = 1018, smaller values are accepted for non-maneuverable subsonic aircraft.

Offset angle(): It is the angle of offset for wheels of MLG relatively to airplane CG.It prevents airplane overturning backward during landing. =  + (12). =11+ 2 = 13.Wheel track(В ) : It is the distance (on front view) between planes of symmetry of MLG wheels. This can be found byB=(0 .15 . .. . 0. 35 )LW

B=0 . 21×22 .39=4 .7 m

Height of airplane center of gravity(Н): It is the distance from airplane CG to the ground.

H= e

tan (γ ) H= 0.462 / tan (13)H= 2.1 m

Height of the landing gear (h): Distance from leg attachment fittings to the runway surface when shock absorber and tiers compression is of parking state (at take-off mass).

[43]

DF = 2.5 m h =H – DF/2h = 2.1 – (2.5/2)h = 0.85 m

Determination of Position of Center of Mass of the Airplane

Position of the airplane center of mass is determined relative to nose part of the wing mean aerodynamic chord (MAC).The recommended distance for the center of mass from the nose part of mean aerodynamic

chord xm as follows:For airplanes with swept wing:

xm=0 . 23×bA=0.23×3. 15=0 .7245(m)Determination of tail armsVertical tail unit arm:It is the distance measured from the aircraft center of massup to the vertical tail unit centre of pressure. It is selected statistically , Tvs = 12m. For T-Tail Tvs is not equal to Ths.Horizontal tail unit arm : It is the distance measured from the aircraft center of mass up to the horizontal tail unit centre of pressure. Horizontal tail unit arm is obtained after drawing theoretical diagram.Calculation of high lifting devices parameters:Flap configuration:

We will consider relative value of flap span from statistical data Lflap=0. 5 to 0. 8

I considered relative length of flap = 0.6The length of the flap is calculated by the formula

Lflap=

Lspan−D fuselage−2 ΔLflap

2×L flap

ΔLflap- is the gap between flap and fuselage = 100mm

Lflap=

22.73−2 . 5−2∗0 .1002

×0 .6=6 m

Flap root chord is calculated by the following formula

b0 flap=b̄ flap⋅b0⋅(1− η−1

η⋅

Dfuselage+2 Δ flap

L )Here

b̄ flap is relative chord of flap is 0.2 to 0.4

[44]

There for b0 flap=0 .2×4 .5⋅(1−4−1

4⋅2.5+2×0 .10022 .73 )=0 . 818 m

Flap tip chord length is calculated by fallowing formula

bкflap=b̄ flap⋅b0⋅(1−η−1

η⋅

Dfuselage+2 Δ flap+2 lflap

L )

bкflap=0 .2⋅¿4 . 5⋅(1−4−1

4⋅2. 5+2×0. 100+2×622 .73 )=0 .46 m

Slats configuration:

We will consider relative value of slat span from statistical data Lslat=0 . 6 to0 . 85

I considered relative length of slat = 0.65The length of the slat span is calculated by below formula

Lslat=

Lspan−D fuselage−2 ΔLslat

2×Lslat

ΔLslat- is the gap between slat and fuselage = 300mmThere for

Lslat=

22 .73−2. 5−2∗0 . 32

×0. 65=6 .3 m

slat root chord is calculated by the fallowing formula

b0 slat=b̄slat⋅b0⋅(1−η−1

η⋅

D fuselage+2 Δslat

L )Here

b̄slat is relative chord of slat is 0.08 to 0.15 = 0.09

There for b0 slat=0 .09×4 .5⋅(1−4−1

4⋅2 . 5+2×0 . 322.73 )=0 .363 m

Slat tip chord length is calculated by fallowing formula

bкslat=b̄slat⋅b0⋅(1− η−1

η⋅

D fuselage+2 Δslat +2l slat

L )

bкslat=0 . 09⋅¿ 4 . 5⋅(1−4−14⋅2 .5+2×0 .3+2×6 . 322.73 )=0 .19 m

Calculation of control surfaces parameters:

Aileron configuration:

[45]

AREA ( SAIL) : It is found by formula SAIL=S AIL×S , from statistical data Relative are of

aileron is 0.06,SAIL=0 .06×63 . 87=3. 83 m2

Length of the aileron is calculated by the following formula

laileron=L−Dfuselage

2−l flap−Δflap−ΔAF−ΔAw

, м,

here Δ AF – the gap between flap and aileron= 0.02

Δ Aw – the gap between aileron and wing tip = 1.2

laileron=22 .73−2. 5

2−6−0 .10−0 . 02−1.2=2.7

m

Aileron root chordbоaileron=b̄aileron⋅b0⋅(1−η−1

η⋅Z̄оaileron)

, м,

Here b̄aileron0.25…0.3 – from the statistical data = 0.25

Here Z̄оaileron=

D fuselage+2⋅(l flap+ΔAf+ΔAw )L .

Z̄оaileron=

2.5+2⋅(6+0 .02+1 .2 )22 .73

=0.74

There for bоaileron=0 .25×4 .5⋅(1−4−1

4×0 .74)=0 .50 m

Aileron tip chord bкaileron=b̄aileron⋅b0⋅(1−η−1

η⋅Z̄кaileron)

, м,

Here Z̄кaileron=

D fuselage+2⋅(lflap+laileron+Δaf +ΔAw)L .

Z̄кaileron=

2 .5+2⋅(6+2 .7+0 .02+1.2 )22.73

=0 .98

There for bкaileron=0 .25×4 . 5⋅(1−4−1

4×0.98)=0.29 m

Elevator configuration: Length of the elevator

lelevator=lH .S

2−ΔHT−ΔHt

, м,Here Δ HT – is the operating gap of elevator =0.02m

[46]

Δ Ht – is the gap between elevator stabilizer to horizontal stabilizer tip=0.6m

lelevator=

7 .552−0 . 02−0. 6=3 .155 m

Elevator root chord

b0 er= b̄er⋅b0 H . S⋅(1−ηH .S−1

ηH . S

⋅2 Δ HTlH .S

), м,

Here b̄er 0.25…0.35 – relative length of elevator chord = 0.25

b0 er=0.25×2.48⋅(1−2−1

2⋅2×0.02

7 .55 )=0 .617 m

Elevator tip chord:

bк er=b̄er⋅b0 H . s⋅(1−ηH ,S−1

ηH .S

⋅lH .S−2 Δ Ht

lH .S), m.

bк er=0 .25×2. 48⋅(1−2−12⋅7 .55−2×0 .6

7 .55 )=0. 359 m

ELEVATOR AREA

S̄EL=S EL/ SHS=0,2…0,4 (lower values – for supersonic aircraft);

SEL= S̄EL × SHS = 0.3 × 13.98 m2 = 4.194 m2

Rudder configuration:rudder area

S̄RUD=0 , 20…0 , 45 (lower values – for supersonic aircraft);

SRD = S̄RD × SVS = 0.40 × 9.01m2 = 3.64 m2

Length of the rudder

lrudder=lV . S−Δ Rf−Δ RT , м,Here Δ RF – the gap between fuselage to rudder root chord= 0.015m

Δ RT – the gap between vertical tail tip to rudder tip chord.=0.6m

lrudder=3 .5−0 .015−0 . 6=2. 885 m

Rudder root chord

[47]

b0 rudder= b̄rudder⋅b0 V .S⋅(1−ηV . S−1

ηV .S

⋅Δ RFlV .S

), м,

Here b̄rudder 0.25…0.4 – relative chord of rudder = 0.25

b0 rudder=0. 25×2.57⋅(1−1−11⋅0 . 02

9 .56 )=0 . 642m

For T-tail tip and root chord of rudder is same there for bkrudder= 0.642 m

SELECTION AND GROUNDS OF AIRCRAFT CONFIGURATION

AERODYNAMIC CONFIGURATION

The “normal" classical configuration applies to my aircraft. The advantages of this configuration are:

wing is in the pure, undisturbed airflow and is not shadowed by stabilizers; nose section of a fuselage is short and does not create destabilizing moment

relatively to the vertical axis; this allows to reduce area and mass of vertical stabilizer;

Crew has better observation of the front semi-sphere.

Selection of wing position relatively to the fuselage

Low-wing aircraft. Advantages: due to ground shield effect (aerodrome surface) Ycr increases, Vto, Vland; decreases height of the landing gear struts and their mass is less, their retraction becomes

simpler; high-lift devices can also be located on ventral wing parts; Safety of passengers and crew increases during emergency landing – the wing

provides additional protection; Floating capacities during emergency landing on water are higher, that allows to

evacuate passengers and crew.

[48]

SELECTION OF WING EXTERIOR SHAPE

Swept wings are applied at M = 0,82.With increase of sweep angle:

the shockwave drag on moderate subsonic and supersonic speeds (Fig. 2.6) is drastically decreased.

M cr=М cr χ=0

21+cos χ

is increased.

Critical values of flutter speed Vfl increase, divergence speed Vdiv (at swept wings), lateral stability increases.

WING SHAPE ON FRONT VIEW

It is characterized by wing dihedral angle

[49]

Dihedral angle is defined by angle between wing chords plane and plane perpendicular to aircraft of symmetry plane passing through the inboard chord. The following types are distinguished:At = 0+7 dihedral angle (for straight wings)

SHAPES OF WING CROSS-SECTIONS AND TAIL UNIT STABILIZER CROSS SECTION

Double convex asymmetrical - high Cy max, smaller Cxр, stable position of the centre of pressure. These airfoils find wide application in various subsonic aircraft;The airfoil should have low profile drag in a range of Cy factors characteristic for cruise flight;

It is necessary that the airfoil with the extended flap has small Cxp at Cmax, especially during climb;

Tip wing cross-sections at Cmax should have smooth performances of shock stall; Internal wing cross-sections should have high values of Cmax with extended flaps; It is necessary to ensure a high value of Mcr above 0,65;

Airfoils for my aircraft are selected as mentioned below

The NACA four-digit wing sections define the profile by

1. One digit describing maximum camber as percentage of the chord.2. One digit describing the distance of maximum camber from the airfoil leading edge in

tens of percent's of the chord.3. Two digits describing maximum thickness of the airfoil as percent of the chord.

[50]

WING AIRFOIL - NACA 2415 airfoil has a maximum camber of 2% located 40% (0.4 chords) from the leading edge with a maximum thickness of 15% of the chord. Four-digit series airfoils by default have maximum thickness at 30% of the chord (0.3 chords) from the leading edge.

HORIZONTAL STABILIZER AIRFOIL - NACA 2412 airfoil has a maximum camber of 2% located 40% (0.4 chords) from the leading edge with a maximum thickness of 12% of the chord. Four-digit series airfoils by default have maximum thickness at 30% of the chord (0.3 chords) from the leading edge.

Selection of the scheme of ailerons

b̄ail=bail /b=0 ,25…0,3; δail .upw=20…25° ; S̄ail=Sail /S=0 ,03…0 , 08 ;

L̄ail=Lail /L=0,2…0,4 ; δail . downw=10…15 ° .

[51]

Requirements to ailerons: Minimum yawing moment (an aircraft yaw motion relatively to OY axis) in bank, with

aircraft turning in the side of bank angle. Full weight balancing with least weight of balance weights. Provision of efficiency on all flight phases. Critical speed of reverse thrust should be sufficient.

SELECTION OF FUSELAGE STRUCTURE

The basic geometrical sizes of the fuselage are selected statistically by comparing the prototypes and are listed belowLf – length of a fuselage; 30.62mDf – diameter of the greatest mid-section, 2.5mSmcs – the area of fuselage mid-section;Lns, length of nose section of a fuselage. 3.75mLts – tail section of a fuselage. 6.25m

n.f – Fitness ratio of fuselage nose section – 1.5

t.f – Fitness ratio of tail section – 2.5

NOSE SECTION

[52]

TAIL SECTIONShapes of nose and tail sections are also determined generally from conditions of

aerodynamics, layout, technology, purpose. For the nose section, an important condition is ensuring the demanded observation from the cockpit that leads to smaller fineness and smaller sharpness. It is reasonable to deflect the tail section of a fuselage upwards to ensure the landing angle during take-off. Many cargo aircraft have a large door in the tail section with the cargo ramp lowered on ground for loading and an unloading of cargos. For modern aircraft, for aerodynamic drag decrease considerations, the whole tail section is lengthened and bended. Some cargo aircraft have the cargo door in the fuselage nose (An-124, C-5А, Boeing 747F). The lower part is capable to open back and upwards that simplifies loading and unloading of transported vehicles and cargos.

CONTOURS OF FUSELAGE NOSE & TAIL SECTION

Yn.f = a (Хn.fdn.f/4n.f) m.

Хn.f – Length of nose section – 3.75m

dn.f – Diameter of fuselage nose section – 2.5m

n.f – Fitness ratio of fuselage nose section – 1.5

m – 0.5 from table

a – 1 from table

Yn.f = 1 (3.75* 2.5 / 4 * 1.5) 0.5 = 1.25Factors m and a are presented in Tab

m 0,35 0,4 0,45 0,5 0,55 0,6 0,65а 1,1293 1,08845 1,04136 1 0,96026 0,9221 0,88546

Y f .ts=± в [ ( Х f . tsd f−Х ) nd f /4 λf . ts ]n.

в – 1 from table

Хt.f – Length of tail section – 6.25 m

[53]

df. – Diameter of the fuselage – 2.5m

t.f – Fitness ratio of tail section – 2.5

n – 0.50 from table

x – Total length of the fuselage – 30.62m

Y f .ts=± 1 [ (6 . 25∗2.5−30 . 62 ) 0 .502 . 5/4∗2 . 5 ]0. 50 = 0.9839

Factors n and в are presented in Table

n 0,30 0,35 0,40 0,45 0,50 0,55 0,60 0,65 0,70

в 18,5396 8,9707 4,3174 2,0728 1 0,481220,2316

20,11147 0,053649

The point of reference of the tail section is located at distance of l f.ns+l f.ts from a fuselage nose. Coordinates of Yf.ns and Yf.ts axis are turned counter-clockwise on n.f and t.f angles. If m and n are selected equal to 0,5 of the contours are usual quadratic parabolas. If smaller values of m and n are selected, shapes of nose and tail sections will be fuller, if greater values are selected, the shapes will be more concentrated.

SELECTION OF STABILIZERS STRUCTURE

T-shaped stabilizers: HS is placed off the zone of wing wash on all flight phases,VS is loaded additionally, its mass is increased

.

DORSAL FIN

[54]

1. Dorsal fin installation (fig. 2.35). The dorsal fin improves VS washing, activates at high speeds, favorably influences side-slip, increases the effective area of VS. Allows to reduce SVS, mVS, VS.

Selection of Landing Gear StructureThe landing gear must ensure: Absorption and dispersion of kinetic energy of impact at landing and taxiing on

aerodrome roughness’s protecting the airframe construction from destruction; Stability and controllability of the aircraft at take-off run, landing roll, taxiing,

maneuvering, pushback;

The three-strut with nose strut configuration: Absence of danger of nose-over (overturning onto the canopy); Capability of effective braking of wheels after landing; Simplification of piloting, preference of landing on two struts with lowering on the

nose strut (the angle of attack decreases and bouncing is excluded); Good directional stability at take-off run and landing roll: the turn force is

compensated by the stabilizing moment of main struts friction forces; Convenience and comfort for passengers, good observation for crew; Simplicity of loading and unloading of cargo aircraft owing to horizontal or nearly

horizontal position of the fuselage axis.Selection of Power Plant Type

It is necessary to select the type of engine by comparing our prototype, two of the engines are mentioned below.

Embraer ERJ 145 - Rolls-Royce AE 3007-A1

[55]

Bombardier CRJ200 - GE CF34-3B1

Then we must select the type of Engine followed by its units and systems and then the integral components of the power plant

ENGINE TYPE - Low Bypass turbojet engines with various bypass ratio without the afterburner.

Integral components of the power plants

1. Engine nacelles with air intakes;

[56]

The intake length in front of the engine face = 1.12 × Engine face diameter = 1.12 × 0.825 = 0.931 m

The maximum nacelle diameter = 1.5 × Engine face diameter

= 1.5× 0.825 = 1.2375 m

The exhaust jet-pipe length aft of the last stage turbine disc = 1.5 × engine face diameter

= 1.5 ×0.825 = 1.05 m

The total length of the Nacelle = Engine length + (k × Engine face diameter)

= 2.78 + (1.5 + 0.825) = 4.018 m

2. Systems of engines attachment;

[57]

3. Fuel and lubrication systems;

4. Fire protection and anti-icing systems;

5. Fuel and oil tanks;

POSITION OR LOCATION OF ENGINE

Position of engines on fuselage tail section allows ensuring the following features: "clean" wing, selection of anhedral/dihedral angle for wing from conditions of optimal lateral stability and controllability of the aircraft, noise level decrease in passenger cabins, fire safety improvement, and engine protection against getting of foreign objects. However, the mass of fuselage tail section increases, the wing mass is increased because of the absence of its unloading, the aircraft center of masses is displaced back, the efficiency of stabilizers decreases, etc. Therefore use of such structure is limited.

SELECTION OF THE SCHEME OF ACCOMMODATION OF CREW AND PAYLOAD

The cockpit or service compartment is usually placed in the forward nose section of the fuselage. Its dimensions depend on crew.Number of crew members is established by calculation: ncrew = 1 2 people for regional airlines; ncrew= 2 people for short-range airways;

Range of my aircraft ShortNo of Recommended crew members Pilot and A Co-pilot

[58]

The number of flight attendants is determined by the following regulations: For I class cabin – one flight attendant for 15-16 passengers; For II and III class cabins– one flight attendant for 30-35 passengers.

Number of cabin attendant = 2

SELECTION OF ENGINE

It is necessary to determine starting thrust of the engine. It is determined on the basis of the

graphical values of starting thrust-to-weight ratio which is determined in lab 4 . For this, it is

necessary to establish the value t0 for the projected airplane.

[59]

LAB 4 : In this part we are finding the Thrust to weight ratio with respect to the aspect ratio of wing taken as 4 and aircrafts wing loading 600 N/m2.

RESULT: In table P, denotes wing loading, Tk, denotes aspect ratio TOW- is the Thrust to weight ratio at take-off which is 0.251

TOB- is the Thrust to weight ratio at landing which is 0.271TOK- is the Thrust to weight ratio at cruising which is 0.179

Engine Total Thrust Determination

P0=t0⋅m0⋅g

[60]

Where t0=0 .251, m0 = 39110 kg

P0=0 . 251×39110×9 . 8=96202(N )P0=96 . 2(kN )

Thrust of one engineFurther starting thrust of one engine can be determined on the basis of engines

numbers “n ”. Number of engines is taken to be 2

Peng=P0

neng

=962022

=48101(N )=48.1(kN )

After determining the total thrust the specific type of engine is selected according to the

thrust value. From engines catalogue based on thrust 48 kN the following engine is selected

ROLLS-ROYCE [Spey RSP 4 Mk 510-14]

Type Low bypass

turbofan

Number of spools : 2

Length 2.8 m Compressor : Axial flow

Fan

Diameter

0.8255 m High pressure compressor

stages :

12

Dry weight 1048 kg Combustors : 10 can annular combustion

chamber

Width 0.92 m High pressure turbine stages : 2

Max 0.942 m Low pressure turbine stages : 2

[61]

diameter

PERFORMANCE DATA

Maximum thrust 48.9 kN Thrust at cruising 30.70 kN

Specific fuel consumption

0.6 Specific fuel consumption atcruising

0.770

By pass ratio 0.71:1 Cruising speed 0.77

Thrust to weight ratio Cruising altitude 32000

Fan pressure ratio 2.70 Airflow 206 lb/s

[62]

DETERMINATION OF CENTER OF GRAVITY OF THE AIRCRAFT

The centre of gravity for the aircraft in preliminary stage is found statistically but before finding it, we need to know the masses of various components. Therefore the mass of the different groups are found by three methods, the first one is the graphical determination using the software which is done already, The second one is the statistical method which is also already done, the last method is the semi-empirical methods

SEMI EMPIRICAL MASS DETERMINATION-FUSELAGE MASS

MFcivil – Cfus × ke ×kp×kuc×kdoor×( 2× L × D× VD0.5)1.5×(MTOM×Nut)x

Where, cfus is a generalized constant to fit the regression, as follows:

0.038 for small unpressurized aircraft (leaving the engine bulkhead forward) 0.041 for a small transport aircraft (≤19 passengers) 0.04 for 20 to 100 passengers 0.039 for a midsized aircraft 0.0385 for a large aircraft 0.04 for a double-decked fuselage 0.037 for an unpressurized, rectangular-section fuselage

All k-values are 1 unless otherwise specified for the configuration, as follows:

ke = for fuselage-mounted engines = 1.05 to 1.07

kp = for pressurization = 1.08 up to 40,000-ft operational altitude1.09 above 40,000-ft operational altitude

kuc = 1.04 for a fixed undercarriage on the fuselage 1.06 for wheels in the fuselage recess

[63]

1.08 for a fuselage-mounted undercarriage without a bulge 1.1 for a fuselage-mounted undercarriage with a bulge

kVD = 1.0 for low-speed aircraft below Mach 0.31.02 for aircraft speed 0.3 < Mach < 0.61.03 to 1.05 for all other high-subsonic aircraft

kdoor = 1.1 for a rear-loading door

For the designed aircraft,VD - Velocity 750km/h = 208 m/s(MTOM×Nut)x 1 for civil aircraftCfus 0.04 for 20 to 100 passenger aircraftke 1.05kp 1.08kuc 1.1kdoor 1.1L - length of the aircraft 30.62mD diameter of the fuselage 2.5mMTOM- maximum takeoff mass 39.11tons

MFcivil = 0.04 × 1.05 ×1.08×1.1×1.1×( 2 30.62× 2.5 × 2080.5)1.5×1 = 5693.969 kg

5% reduction in mass due to composite usage therefore 5409.270kg

WING MASS

M w=C w ×k uc ×k se× k sp× k wl× k ℜ× Sw0.78× AR ×(1+λ)0.4 ×(1− fuel mass

MTOM)

0.4

cos Λ ( tc )

0.4

where Cw = 0.0215 and flaps are a standard fitment to the wing.

kuc = 1.002 for a wing-mounted undercarriage; otherwise, 1.0ksl = 1.004 for the use of a slatksp = 1.001 for a spoilerkwl = 1.002 for a winglet (a generalized approach for a standard size)kre = 1 for no engine, 0.98 for two engines, and 0.95 for four engines (generalized)

[64]

CW 0.0215kuc 1.002kse 1.004ksp 1.001kwl 1.002kre 0.98Sw - Wing area 63.87m2

AR – Aspect ratio 7.35λ 0.375Fuel mass 8760kgΛ - Sweep angle 22.73

( tc ) - Thickness ratio

0.18

M w=0.0215 × 1.002× 1.004 ×1.001 ×1.002 ×0.98 ×63.870.78× 7.35 ×(1+0.375)0.4 ×(1− 8760

39110)

0.4

cos22.73 (0.18 )0.4

MWing = 2682.738kg

HORIZONTAL TAIL MASS

MHT =0.02×(MTOM × N ut )0.48× Sh 0.78 × K mat × AR ×(1+λ)0.4

cos Λ( tc )

0.4

Kconfg 1.1 for T-tail Sh – Area of horizontal stabilizer 13.98 m2

AR – Aspect ratio of the horizontal stabilizer 4Λ – Sweep angle of the of the horizontal stabilizer

20

( tc ) - Thickness ratio of the horizontal

stabilizer

0.12

λ 0.3Kmat 0.98

[65]

MHT =0.02×(39110 ×3.5 ×1.08)0.48× 13.980.78 ×0.98 × 4 ×(1+0.3)0.4

cos20 (0.12 )0.4 ; MHT = 806 kg

VERTICAL TAIL MASS

MVT =0.02× K conf ×(MTOM × N ut)0.48× S v0.78 × K mat × AR×(1+λ)0.4

cos Λ( tc )

0.4

Kconfg 1.1 for T-tail Sv - Area of vertical stabilizer 9.01 m2

AR- Aspect ratio of the vertical stabilizer 1.36Λ - Sweep angle of the of the vertical stabilizer

32

( tc ) - Thickness ratio of the vertical stabilizer

0.10

λ 0.3Kmat 0.98

MVT =0.02× 1.1×(39110× 3.5 ×1.08)0.48× 9.010.78 ×0.98 ×1.36 ×(1+0.3)0.4

cos 32 (0.10 )0.4 ; MVT =274.212 kg

TOTAL TAIL UNIT MASS

MTAIL = Horizontal tail group mass + Vertical tail group mass = 806+ 274.212 = 1080.212 kg

NACELLE MASS

For by-pass ratio less than 4,MNAC = 6.2 × thrust per nacelleMNAC = 6.2 × 48 kN = 297.6 kgFor two nacelles, MNAC = (6.2 × 48) 2 = 595.2 kg

[66]

UNDERCARRIAGE MASS

For Low wing aircraft,MUG = 0.04× MTOM MUG = 0.04× 39110 kg = 1564.4 kg

POWERPLANT MASS

For turbofan ( without thrust reverser )Meng = 1.4 ×Mdry engine (Mdry engine = 1048kg)Meng = 1.4 ×1048 kg = 1467.2 kgFor two engines Meng = 2934.4 kg

SYSTEMS MASS

MSYS = 0.12 × MTOMMSYS = 0.12 × 39110 kg = 4693.2 kg

FURNISHING MASS

MFURN = 0.057 × MTOMMFURN = 0.057 × 39110 kg = 2262.56 kg

CONTINGENCY & MISCELLANOUS

MCONT = 0.015 × MTOMMCONT= 0.015 × 39110 kg = 586.65 kgMMISC = 0.010 × MTOMMMISC= 0.010 × 39110 kg = 391.1 kg

CREW MASS

Crew members = 4MCREW = 4 × 80 = 320 kg

PAYLOAD MASS

[67]

No of passengers = 47Mass of one passenger = 90 kgMPAYLOAD = 47 ×90 = 4230 kg

AVERAGE BETWEEN GRAPHICAL, SEMI-EMPIRICAL & STATISTICAL METHOD

Mass of Airframe = Mfuselage + Mwing + Mtail + Mlg

FUSELAGEMfuselage Graphical method = 13688.5 kgMfuselage Statistical method = 4392.83 kgMfuselage Semi-empirical method = 5409.270 kg

Mfuselage Average = 13688.5+4392.83+5409.270

3=7830.2kg

WING Mwing Graphical method = 1877.28 kgM wing Statistical method = 4956.01 kgM wing Semi-empirical method = 2682.738 kg

M wing Average = 1877.28+4956.01+2682.738

3=31 72.009 kg

TAILMtail Graphical method = 711.802 kgM tail Statistical method = 863.54 kgM tail Semi-empirical method = 1080.212 kg

M tail Average = 711.802+863.54+1080.212

3=8 85.1846 kg

LANDING GEARMlg Graphical method = 2424.82 kgMlg Statistical method = 2302.79 kgM lg Semi-empirical method = 1564.4 kg

Mlg Average = 2424.82+2302.79+1564.4

3=2097.33 kg

[68]

POWERPLANTMpowerplant Graphical method = 3363.46 kgM powerplant Statistical method = 5475.4 kgM powerplant Semi-empirical method = 2934.4 kg

Mlg Average = 3363.46+5475.4+2934.4

3=3924.42 kg

POWERPLANTMfuel Graphical method = 8760.64 kgM fuel Statistical method = 7039.8 kg

M fuel Average = 8760.64+7039.8

2=7900.22kg

EQUIPMENT & CONTROL SYSTEMSMeq & cont Graphical method = 4952 kgM eq & cont Statistical method = 4383.839 kgM eq & cont Semi-empirical method = 4693.2 kg

M eq & cont Average = 4952+4383.839+4693.2

3=4676.3463kg

FURNISHINGMfurnishing Graphical method = 610.116kgM furnishing Statistical method = 2346.6 kg

M furnishing Average = 610.116+2346.6

2=2262.560kg

MASS BREAKDOWNSTRUCTURAL GROUP MASSStructural group mass = M fuselage + M wing + M tail + M lg

M fuselage = 7830.2 kgM wing = 3172.009 kgM tail = M horizontal + M vertical

M tail = 480+ 375.18 = 885.1846 kgM lg = M main lg + M nose lg

M lg = 1400 + 697.33 = 2097.33kg

Structural group mass = 7830.2 + 3172.009 + 885.1846 + 2097.33Structural group mass = 13984.723 kg

POWERPLANT GROUP MASS

[69]

Powerplant group mass = M dry engine + M nacelle + M fuel system+ M oil + M control system

M dry engine = 2096 kgM nacelle = 595.2 kgM fuel system = 650 kgM oil = 383 kgM control system = 200.22 kg

Powerplant group mass = 2096+ 595.2 + 650+ 383+ 200.22Powerplant group mass = 3924.42 kg

FUEL MASS GROUPM fuel = 7900.22

M fuel Group 1 = 3950.11 kg

M fuel Group 2 = 3950.11 kg

AUXILIARY POWER UNITMapu = 7 × Number of seats

Mapu = 7 × 52 = 364 kg

EQUIPMENT & CONTROL SYSTEMS GROUP MASS Equipment & control systems = M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M avionics &

electronics + M operating inst

M ecs = 15 × Number of seatsM ecs = 15 × 52 = 780 kgMfcs = 550 kgM hyd & pneum = 1400 kgM electrical = 13 × Number of seatsM electrical = 13 × 52 = 672 kgM inst & navi = 365 kgM avionics & electronics = 405 kg

M operating inst = 504.34 kg

Equipment & control systems = 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463Equipment & control systems = 4676.3463 kg

FURNISHING GROUP MASS Furnishing group mass = M seat + M paint + M oxygen system

M seat = 15 × Number of seats = 15× 54 = 810 kgMpaint = 338.358 kgM oxygen system = 300 kg

[70]

Furnishing group mass = 810+ 338.358+ 300 = 1449 kg

CONTINGENCY & MISCELLANEOUS GROUP MASSContingency & miscellaneous = M contingency + Mmiscellaneous

M contingency = 586.65 kgMmiscellaneous = 391.1 kgContingency & miscellaneous = 586.65 + 391.1 = 977.75 kg

PAYLOAD, CREW, LUGGAGE, CONSUMABLES & WATERPAYLOAD MASSNo of passengers = 47Mass of one passenger = 75 kgMpayload = 47 ×75 = 3525 kg

CREW MASSCrew members = 4Mcrew = 4 × 80 = 320 kg

LUGGAGE MASSNo of passengers = 47Mass of luggage for one passenger = 15 kgMluggage = 47 ×15 = 705 kg

CONSUMABLES & WATERMConsumables = number of passengers ×10kgMConsumables = 47 ×10kg = 470 kg

MAXIMUM TAKE-OFF MASS

MTOM(maximum takeoff mass) = OEM + Payload + FuelOperating Empty mass (OEM) is the mass of structure, powerplant, furnishing systems, unusable fuel and other unusable propulsion agents, and other items of equipment that are considered anintegral part of a particular airplane configuration. Also included are certain standard items,personnel, equipment, and supplies necessary for full operations, excluding usable fuel and payload.OEM(operator’s empty mass) = MEM + Crew + ConsumableWhere,

[71]

MEM (manufacturer’s empty mass) = M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle + M fuel system+ M oil + M control system + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M

avionics & electronics + M operating inst )+ (M seat + M paint + M oxygen system )+ (M contingency + Mmiscellaneous)

MEM (manufacturer’s empty mass) = 7830.2 + 3172.009 + 885.1846 + 2097.33 + 2096+ 595.2 + 650+ 383+ 200.22+ 364 + 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463+ 810+ 338.358+ 300 + 586.65 + 391.1

MEM (manufacturer’s empty mass) = 25375 kg

OEM(operator’s empty mass) = M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle + M fuel

system+ M oil + M control system + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M avionics &

electronics + M operating inst )+ (M seat + M paint + M oxygen system )+ (M contingency + Mmiscellaneous) +( special equipments )+ Mcrew + MConsumables

OEM(operator’s empty mass) = 7830.2 + 3172.009 + 885.1846 + 2097.33 + 2096+ 595.2 + 650+ 383+ 200.22+ 364 + 672+ 405 + 780+ 365+ 550+ 1400 + 504.3463+ 810+ 338.358+ 300 + 586.65 + 391.1+320 + 470

OEM (operator’s empty mass) = 26165.58 kg

Maximum Design Takeoff Weight (MTOM). Maximum weight for takeoff as limited by aircraftStrength and airworthiness requirements. (This is the maximum weight at start of the takeoff run.)

MTOM = (M fuselage + M wing + M tail + M lg) + (M dry engine + M nacelle + M fuel system+ M oil + M control system

+ M fuel + Mapu) +( M ecs + M fcs + M hyd & pneum+ M electrical+M inst & navi + M avionics & electronics + M operating

inst )+ (M seat + M paint + M oxygen system )+ (M contingency + Mmiscellaneous) + ( special equipments )+ (Mpayload + Mcrew + Mluggage) + MConsumables

MTOM =( 7830.2 + 3172.009 + 885.1846 + 2097.33) + (2096+ 595.2 + 650+ 383+ 200.22)+ 7900.22+ 364 +(672+ 405 + 780+ 365+ 550+ 1400 + 504.3463)+ (810+ 338.358+ 300) + (586.65 + 391.1 )+(814)+ (3525 + 320 + 705) + 470

MTOM = 13984.723+3924.42+7900.22+364+4676.3463+1449+977.75+814 +4550+470

MTOM = 39110 kg

CENTER OF GRAVITY

[72]

From the required mass data the CG of the aircraft is determined by tabulating all the unit cargo with their position located on the aircraft along with its X and Y co-ordinates. Some of the coordinates are assumed based on the statistical data and others can be located on three view diagram. At last the average of the CG is found and presented in the A1 format.

STRUCTURE

UNIT CARGO POSTION OF CG

X LOCATION

Y LOCATION MASS

Wing 0.40 × Bmac0.40 × 3.15m = 1.26m

18.551m 1.431m 3172 kg

Fuselage 0.6×Lf0.6×30.62 = 18.37m

18.372m 2.34m 7830 kg

Horizontal stabilizer

0.45×Bahs0.45×1.93= 0.8685m

30.533m 5.601m 480 kg

Vertical tail 0.45 × Bavs0.45 ×2.57= 1.1565m

28.919m 4.459m 405 kg

Main LG Located aft of CG of aircraft

17.679m 3.75m 1400 kg

Nose LG From 3 view 4.265m 1.322m 697 kg

POWERPLANT

Middle engine Center section 23.54m 2.887m 1400 kgEnd engine Center section 25.772m 2.837m 696 kgEngine nacelles Center section 24.028m 2.887m 595 kgFuel systems+Oil System+

Center of fuel tanks

18m 1.5m 650+

383+

200

[73]

Engine Control system

=1233 kg

EQUIPMENT & CONTROL SYSTEMS

Environmental Control System+Auxilliary power unit

At tail section 25.772m 2.837m 780+

364=

1144 kgNavigation +Flight Control System+Avionics & Electronics

0.40 × nose length0.40 × 3.75=1.5m

1.5 2 365+550+405=

1320 kg

Electrical0.75×Fuselage length0.75 ×30.62=22.965m

22.96 2.1 672 kg

Hydraulics+Operating unit

Near main LG 16.773 1.3 1400+

504=

1904 kgFurnishings+Special equip+Contingency

0.51×Fuselage length0.51×30.62=15.611m

15.611 2.3 1449+

814+

977=

3240 kg

OPERATIONAL ITEMS

Crew 0.45 × nose length 1.6875 2.499 160 kg

[74]

0.45 ×3.75=1.6875

Cabin attendant +Consumables

Near cabin 5 2 160 +470=

630 kgFUEL

Group- 1 Center of fuel tanks

19 1.414 3950 kg

Group- 2 Center of fuel tanks

19 1.414 3950 kg

PAYLOAD

Passenger 0.51×Fuselage length

0.51×30.62=15.611m

15.611m 2.3 3525 kg

Luggage 0.51×Fuselage length

0.51×30.62=15.611m

15.611m 2.3 705 kg

DETERMINATION OF Mg X and Mg YSTRUCTURE Mg Mg X Mg YUNIT CARGO MASS X LOCATION Y

LOCATIONWing 3172 31117.32 577257.4033 44528.88492Fuselage 7830 768123.3 14111961.27 1797408.522Horizontal stabilizer

480 4708.8 143773.7904 26373.988

Vertical tail 405 3973.05 114896.633 11715.82995

Main LG 1400 13734 242803.386 51502.5Nose LG 697 6837.57 29162.236 25640.887

[75]

POWERPLANTMg Mg X Mg YMiddle engine 1400 13734 323298.36 39650.058End engine 696 6827.76 175965.0307 19370.35512Engine nacelles 595 5836.95 140250.2346 16851.27465Fuel systems+Oil System+Engine Control system

650+

383+

200=

1233

12095.73 217723.14 18143.595

EQUIPMENT & CONTROL SYSTEMS Mg Mg X Mg Y Environmental Control System+Auxilliary power unit

780+364=1144

11222.64 289229.8781 31838.62968

Navigation +Flight Control System+Avionics & Electronics

365+550+405 =1320

12949.2 19423.8 25898.4

Electrical672 6592.32 151359.6672 13843.872

Hydraulics+Operating unit

1400+504=1904

18678.24 313290.1195 24281.712

Furnishings+Special equip+

1449+814+977

31784.4 496186.2684 73104.12

[76]

Contingency =3240

OPERATIONAL ITEMS

Crew 160 1569.6 2648.7 3922.4304Cabin attendant +Consumables

160 +470=630

6180.3 30901.5 12360.6

FUEL

Group- 1 3950 38749.5 736240.5 54791.793

Group- 2 3950 38749.5 736240.5 54791.793

PAYLOAD

Passenger 3525 34580.25 539832.2828 79534.575Luggage 705 6916.05 107966.4566 15906.915

∑i=1

n

(mg)i = 1074960.48

∑i=1

n

(mgx )i = 19500411.16

∑i=1

n

(mgy )i = 2440660.734

CG coordinates.

xm=∑i=1

n

(mgx )i/∑i=1

n

(mg )i

Xm = 19500411.161074960.48

=18.140

ym=∑i=1

n

(mgy )i/∑i=1

n

(mg)i.

[77]

Ym = 2440660.731074960.48

=2.2

DESIGN STRUCTURAL CONFIGURATIONDESIGN STRUCTURAL CONFIGURATION OF WING

The choice of load-carrying structures for wing is determined by

criterion of bending moment intensity acting upon wing,

M

H3, MPa;

criterion of wing specific load, p0 = m0g/s, N/m2; wing lay-out (arrangement of fuel tanks or fuel sections, wells for LG

retraction, hatches for instrument and equipment maintenance and so on inside the wing);

fuselage lay-out and possibility to arrange the wing center section inside the fuselage;

Requirements of sufficient strength, rigidity, aeroelasticity, service life, reliability under condition of minimal mass, costs.

Wing LCS is selected on the basis of the following factors:

To define approximately the wing LCS let us use concept of conditional spar, which caps have width bcond. about 0.6b, here b – wing chord in design section

[78]

Conditional sparbcond = 0.6 × Chord of wing at design sectionbcond = 0.6 × 3.82 = 2.292Nbr. cap = bcond. cond. br.Nbr. cap = 2.292 ×2×10-03.×330×106Nbr. cap = 1.512 Mpa

CAP THICKNESS OF CONDITIONAL SPAR Depending upon the Cap thickness of spar of the wing the distance between Ribs and Stringers are selected, from the formula the cap thickness is determined as follows,

р0 – wing specific load, N/m2; 600×9.81 = 5886 N/m2

S – wing area, м2; 63.87 m2

g – gravity acceleration, 9.8 m/s2; 9.8zA – MAC coordinate from longitudinal airplane axis spanwise, m;

4.4m

mi, zi – mass of cargo, placed on wing, kg, and its CG coordinate, m;

[ mass of fuel×Zfuel+ mass of LG×ZLg

]9.81[7900 × 5.5 +1400 ×1.8]

mwing – wing mass, kg; 3172.009 kgndes. – design g-force, for design case А it equals 1213.5 for maneuverable aircraft, it equals 69 for limited maneuverable aircraft and 2.53.8 for non-maneuverable passenger and transport;

3

[79]

c̄ – wing thickness ratio; 0.15 [NACA 2415]

b0 – wing root chord, m; 4.5 mbr. – breaking stress of spar capbr = 330 МPа for cap made of aluminum alloy Д16Т, 880 МPа for cap made of alloyed steel 30ХГСА, 800 МPа for titanium alloy Вт6;

330 MPa

= 2mm

cond. = 2 mm defines class of spar-type wings with partially working skin. Bending moment is taken by spar; cross-cut force Q is taken by spar webs, torsion torque is taken by contour shaped by skin and spar webs. To increase skin critical stress in shear it is reinforced by weak stringers and thickly arranged ribs.

Recommended Rib pitch 220 mm

Stringer pitch 200 mm

CRITERION OF LOAD MOMENT INTENSITY AND CROSS-CUT FORCELOAD MOMENT INTENSITYAfter finding the rib and stringer pitch, we need to find the type of wing load carrying structure. This can be found using load moment intensity formula and from its result it is desirable to select appropriate LCS.

MH3=

((5886×63 . 87−3172 . 009×9. 8 )×4 . 4−2×24696×1.8×9 .8−2×7900.22×5 . 5×9 .8 )×30. 31677

M

H3 =5.8369 MPa

Since the intensity of bending moment does not exceed 15 МPа, it is better to use spar-type wing.

SPAR-TYPE WING CONSTRUCTION: - we are considering two cross-sections of the wing, which are, the root chord and the second one is wing tip chord section.

Root chord 4.5 m

M

H3=(( p0 S−mw⋅g )⋅z A−2mi⋅g . zi )⋅n p

1 .03⋅(c⋅b0 )3

δ ln=[ 168616 .8−2(425810−24696 )−136777 . 0281 ]3

0 .96×0 . 15×(4 . 52)⋅330⋅106=0 .00204 (m)

[80]

Tip chord 1.1 m

Spar position in the wing:-

Front spar Rear sparAt root chord = 0.25 × 4.5 = 1.125 m At root chord = 0.74 × 4.5 = 3.33 mAt tip chord = 0.25 × 1.1 = 0.275 m At tip chord = 0.74 × 1.1 = 0.814 m

CROSS-CUT FORCE

To calculate intensity of the CROSS-CUT FORCE the concept of equivalent rectangular wing is used go by turning swept wing so to provide perpendicularity of line of elastic centers to airplane longitudinal axis. For straight wing such turn is not accomplished. Position of line of elastic centers may be accepted at 40% of chord.

[81]

The formula for cross cut force is,

Q – cross-cut force;Н – design height of wing section airfoil, m;

zcut. – distance from design section to point of application of resultat force of aerodynamic and mass loads, m; =

9.703 [ 3.8+2(1.1)

3.8+1.1 ] = 3.96m

Scut. – area of wing cut part; 23.866 m2

mwing – mass of wing; 3172 kgLcut – length of wing cut part; 9.72 mm0 – airplane take-off mass; 39110mi – mass of concentrated cargo (mass of fuel + mass of LG) 9.81nop × f 3

c̄ – wing thickness ratio; 0.15 [NACA 2415]

b0 – wing root chord, m; 4.5 m

QΣ

H = 0.9493 MPa

Position and Purpose of Strong Ribs in the wing

Number of the Strong Rib

Purpose of the Strong Rib

Fuselage ribFront Spar Joint in the WingRear Spar Joint in the WingMain landing gear mechanism supportFlap Joint Main landing gear mechanism supportMain Landing Gear, Flap Joint

[82]

Engine Connection, Flap JointEngine Connection, Flap Joint Flap JointFlap Joint Flap JointAileron JointAileron JointAileron JointAileron JointAileron Joint Aileron Joint

LOAD-CARRYING STRUCTURE OF THE FUSELAGESEMI-MONOCOQUE-Advantages

1) The semi-monocoque structure of a fuselage ensures taking up and transmission of all loads at high stiffness, strength and low mass. The reason for this is that the material, which carries bending and torsion stresses (skin and stringers), is as much possible distance from the central axis.

2) Semi-monocoque has large internal free volumes and allows them to be effectively used.

The distance between frames depends on thickness of a fuselage skin, configuration and mass of the airplane. On real structures frame pitch is accepted is some limits, for Light transport airplanes it is between 300 to 400mm. The distance between stringers in a fuselage is chosen for the same reasons as in a wing. Depending on thickness of a skin distance between stringers is accepted as 80…250mm for medium and light airplanes.

Distance between frames = 400mm.Distance between stringers = 150mm.

Position and Purpose of Strong Ribs and Frames in Aircraft StructureOn Fuselage:Number of the Frame

Purpose of the Frame

Nose Radom SupportConnection with the Cock-PitMain Door support beam connection

[83]

i) Landing gear cutout support connectionii) Main Door Connectioni) Connection between Nose and mid fuselage sections.ii) Landing gear cutout support connection

Wing front spar connectionEmergency door support beam connectionEmergency door support beam connectionSecond emergency exit support beam connection

i) Wing rear spar Connectionii) Emergency door support beam connection

Connection between fuselage mid section and rear sectionRear Door ConnectionRear Door ConnectionVertical Tail front Spar Connection to the FuselageVertical Tail rear Spar Connection to the Fuselage

Load-Carrying Structures for the Tail Unit SectionStructural members of horizontal and vertical must be coordinated with each other

and with structural members of the fuselage also. In the design of tail unit, the two-spar structure is usually applied. The lightest structure of tail unit can be in that case if the primary structural members of the tail unit carrying bending stresses (spar and torsion-box) may be passed through the fuselage. To reduce structure mass whenever possible vertical unit and stabilizer should be fixed to the same strong frames of the fuselage. Load-Carrying Structures of Vertical Tail UnitSpar-type vertical tail unit construction: - we are considering two cross-sections of the wing, which are, the root chord and the second one is wing tip chord section. Root chord 2.57 mTip chord 2.57 m

SPAR POSITION Front spar Rear sparAt root chord = 0.18 × 2.57 = 0.462 m At root chord = 0.63 × 2.57 = 1.6191 mAt tip chord = 0.18 × 2.57 = 0.462 m At tip chord = 0.63 × 2.57 = 1.6191 m

Purpose of Strong RibsNumber of the Strong Rib

Purpose of the Strong Rib

Fuselage rib

[84]

Nose Spar JointStarting Connection Rear Spar JointRudder JointRudder JointRudder Joint

Load-Carrying Structures of Horizontal Tail UnitSpar-type Horizontal Tail Unit construction: - we are considering two cross-sections of the wing, which are, the root chord and the second one is wing tip chord section.

Root chord 2.48 mTip chord 1.24 m

Spar position Front spar Rear sparAt root chord = 0.20 × 2.48 = 0.496m At root chord = 0.65 × 2.48 = 1.612 mAt tip chord = 0.20 × 1.24 = 0.248 m At tip chord = 0.65 × 1.24 = 0.806 m

Purpose of Strong RibsNumber of the Strong Rib

Purpose of the Strong Rib

Fuselage Starting Connection of the Horizontal TailNose spar connectionRear spar connectionElevator Joint Elevator JointElevator Joint

LANDING GEAR

In aviation, the under carriage  or landing gear is the structure (usually wheels, but sometimes skids, floats or other elements) that supports an aircraft on the ground and allows it to taxi, takeoff and land. Landing gear usually includes wheels equipped with shock absorbers for solid ground, but some aircraft are equipped with skis for snow or floats for water. To decrease drag in flight some undercarriages retract into the wings and/or fuselage with wheels flush against the surface or concealed behind doors; this is called retractable

[85]

gear. In this project work, the aircraft is equipped with such type of landing gear configuration. Here tricycle type landing gear with nose /front support is chosen, as in most of the passenger pla

STANDARD SPECIFICATION OF THE DESIGNED AIRCAFTMAERO – DA12

The designed aircraft is the all-metal low-wing cantilever monoplane with Т-tail with fixed fuselage-mounted stabilizer.

The aircraft is a regional passenger aircraft, designed to be operated on domestic airlines for transporting passenger’s patients and injured man at short- and medium-haul service airlines.

Two turbofan Rolls Royce [ spey RSP 4 Mk 510-15] (2х10790 kgf) and RE220 gas turbine engine of the auxiliary power unit are installed in the aircraft.Aircraft crew consists of tree members:

- captain (pilot);- co-pilot;

PERFORMANCESThe maximum allowed operational enroute speed and Mach number for enroute aircraft configuration are the following:

VmaxOp 835 km/hМmaxOp 0.65

DESIGN MASSES - Maximum takeoff mass 39110 kg- Empty mass 25375 kg

CENTER OF GRAVITY DATAAllowable range of center of gravity at takeoff, in flight and at landing:

-- extreme forward 15.5 % MAC-- extreme aft 31.5 % MAC

DIMENSIONS AMD AREASWing:

- area 63.87 m 2

- span 22.39m- inner wing sweep along line of 0.25 chord 26° 34'- ailerons' area 3.83 m 2

[86]

Tail unit:- horizontal tail area 13.98 m2

- elevator area 4.194 m2

- vertical tail area 9.01 m2

- rudder area 3.64 m2

Fuselage:- length 30.62 m- maximum diameter 2.5 m

Inner dimensions of Passenger compartment:- length 18.03 m- width along floor 2.08 m- maximum width 2.27 m- height 1.86 m

Entrance door dimensions 0.91x1.78 mUpper emergency hatch dimensions 0.480.510 mGalley service door dimensions 0.61 1.22m Over wing emergency exit 0.51 0.97m Landing gear:

- track 4 .7 m- wheelbase 14.6976 m

Wheels' parameters:а) main landing gear:

- tire dimensions 800х310 mm- pressure 0.86 МPа

b) nose landing gear:- tire dimensions 500х310 mm- pressure 0.86 МPа

Conclusion: In the general designing part we calculated the required geometrical parameters of the designed aircraft, the components mass and the three views is sketched finally according to the results.

[87]

DESIGN SECTION PART 2

2. CALCULATION OF AERODYNAMIC CHARACTERISTICS OF THE AIRCRAFTPARAMETERS OF AEROPLANE

PARAMETER ABBREVATION DIMENSION UNITtake off mass M t 0 39110 kg

number of passengers N pass 47

maximum speed V MAX 760 km/h

cruising speed V cr 835 km/h

range LH 2500 km

engine thrust P° 48 kn

GEOMETRICAL CHARECTERISTICS OF FUSELAGE

PARAMETERS ABBREVATION DIMENSION UNITfuselage length LF 30.62 M

length of fuselage nose LN 3.75 M

length of fuselage tail LT 6.25 M

length of mid fuselage LMF 20.62 M

diameter of fuselage DF 2.5 M

nose deflected angle βnose 3 deg

tail deflected angle β tail 4 deg

aspect ratio of fuselage λF 12.25 -

aspect ratio of fuselage nose λN 1.5 -

aspect ratio of fuselage of tail λT 2.5 -

[88]

CHARACTERISTICS OF WING WITH VENTRAL SECTION

PARAMETERS ABBREVATION DIMENSION UNITwing span LW 22.39 M

gross area SW 63.87 M 2

root chord b0 W 4.5 M

tip chord bkW 1.1 M

mean aerodynamic chord b AW 3.15 M

swept angle X 0.25 22.73 DEG

aspect ratio λW 7.85 -

taper ratio ηW 4 -

fuselage diameter D 2.5 M

thickness ratio C⃗ 0.18 -camber ratio f⃗ 0.02 -

position of chamber relative x⃗ f 0.35 -

AERODYNAMIC CHARECTERISTICS OF WINGDetermination of angle of Zero lifts

α °0 = -60 f⃗ [1+10 (x⃗ f−0.2) 2]

α °0 = -60x 0.02[1+10 (0.35−0.2) 2] = -1.47

[89]

Determination of derivative of lift coefficient

c yα

∞= 2 π ( 1- 0.274 √c )c y

α∞= 2 × 3.14(1- 0.274 √0.18 ) = 5.1781

m= λ

c ya∞α . cos x0.25

m= 7.85

5.1781.cos (22.73)= 1.64

τ=0.17√m[(ηw )2+ 1

(5ηw+1)3 ]

τ=0.17√1.64 [(0.244 )2+ 1

(5 ×0.244+1)3] = 0.03292

c yα =

πλ

1+τ+√(1+τ)2+¿¿¿

c yα =

π 7.85

1+0.03292+√(1+0.03292)2+¿¿¿ = 3.9149

Determination of maximum coefficient of lift

Re = v .ba

v

= 835 × .3 .15

3.6 x1.46 x10−5 = 500.42 × 105

c ya max∞ ≈39.3 c⃗ exponent−8 c⃗ . tg (Re . 10-6)c ya max∞ ≈39.3 0.18 exponent−8 x 0.18 . tg (500.42 × 105 . 10-6) = 1.6760

c yamax = c yamax ∞(1-

0.49 n+0.19n+1

sin2 x0.25)

c yamax= 1-1.6760¿] = 1.5327

α cr° = 57.3 ×

c yamax

c y aα

+ α °0 + 1.5°

α cr° = 57.3 ×

1.53273.9149

+ (-1.47) + 1.5° = 22.463

[90]

Drag coefficient

A =1+∂π . λω

A =1+0.03

3.14 .7.85 = 0.04178

C x 0friction = 2C f ηe ηm

C f = 0.087¿¿¿ = 0.087

¿¿¿ = 2.338×10−3

ηm= 1+5¿¿ = 1+5¿¿ = 1.0621

ηe= 1+2(τ)+9 (τ )1+2(τ )+9(τ)2

= 1+2(0.18)+9(0.18)1+0.36+9(0.18)2

= 1.6516

C x 0friction = 2C f ηe ηm =2×2.338 × 10−3 × 1.0621×1.6516=8.2024×10−3

Aerodynamic quality

Kmax = 1

2√A C x 0 fr =

1

2√0.04178 . 8.2024 × 10−3 = 27.0094

C ya Kmax = √C x0

A = √ 8.2024 ×10−3

0.04178 = 0.44308

C yaline = C yaα (α−α0) = 3.9149(9°−(−1.47°)) = 0.714860

C xa = C x 0 + AC yaline2 = 8.2024×10−3+ 0.04178(0.714860)2= 0.026

K=C ya

C xa =

0.7148600.026

= 27.4946

[91]

Determination of Aerodynamic Center

M Z 0 ≈-2.8 f⃗ (1-√1−2 x⃗ f )M Z 0 ≈-2.8 x 0.021 ( 1-√1−2 x0.35 ) = -0.02532

M ZCyα = -0.25(1-1.6c⃗2)

M ZCyα = -0.25(1-1.6x0.182) = -0.2370

M Z = [cya (M ZCyα) +M Z 0]

M Z =c ya(0.23)-0.0253M Z = [0.71486(-0.23) + (-0.02532)] = -0.194741

CHARECTERISTICS OF HORIZONTAL STABILIZER WITH VENTRAL SECTION

PARAMETERS ABBREVATION DIMENSION UNITspan of horizontal stabilizer LHS 7.55 M

gross area SHS 13.98 M 2

root chord bOHS 2.48 M

mean aerodynamic chord b AHS 1.93 M

sweep angle xO.25 HS 20 DEG

aspect ratio λHS 4.077 -

taper ratio ηHS 2 -

tip chord bKHS 1.24 M

thickness ratio CHS 0.12 -

Determination of angle of Zero liftsα °

0 = -60 f⃗ [1+10 (x⃗ f−0.2)2]α °

0 = -60x 0.02[1+10 (0.4−0.2)2] = -1.68

[92]

Determination of derivative of lift coefficientc y

α∞= 2 π ( 1- 0.274 √c )

c yα

∞= 2 × 3.14( 1- 0.274 √0.12 ) = 5.54

m= λ

c ya∞α . cos x0.25

m= 4.077

5.549. cos (20)= 0.78188

τ=0.17√m [(ηw )2+ 1

(5ηw+1)3 ]

τ=0.17√0.7818 [(0.5 )2+ 1

(5 × 0.5+1)3] = 0.041080

c yα =

πλ

1+τ+√(1+τ)2+¿¿¿

c yα =

3.14 x 4.077

1+0.0410+√(1+0.0410)2+¿¿¿ = 3.452

Determination of maximum coefficient of lift

Re = v .ba

v

= 835 × .1 .93

3.46 x1.46 x10−5 = 306.6114 × 105

c ya max∞ ≈39.3 c⃗ expone nt−8 c⃗ . tg (Re . 10-6)c ya max∞ ≈39.3 .18 exponent−8 x 0.18 . tg (306.6114× 105 . 10-6) = 1.6760

c yamax = c yamax ∞(1-

0.49 n+0.19n+1

sin2 x0.25)

c yamax= 1-1.6760¿] = 1.79

α cr° = 57.3 ×

c yamax

c y aα

+ α °0 + 1.5°

α cr° = 29.68

[93]

Drag coefficient

A =1+∂

π . λHS

A =1+0.03

3.14 .4.077 = 0.0804

C x 0friction = 2C f ηe ηm

C f = 0.087¿¿¿ = 0.087

¿¿¿ = 2.5106×10−3

ηm= 1+5¿¿ = 1+5¿¿ = 1.0621ηe= 1+2(τ)+9 (τ ) = 1+2(0.12)+9(0.12) =1.3696

C x 0friction = 2C f ηe ηm =2×2.5106 × 10−3 × 1.0621×1.3696=7.3040×10−3

Aerodynamic quality

Kmax = 1

2√A C x 0 fr =

1

2√0.0804 .7.3040 ×10−3 = 20.63296

C ya Kmax = √C x0

A = √ 7.3040 × 10−3

0.0804 = 0.30140

C yaline = C yaα (α−α0) = 3.4526(9°−(−1.47°)) = 0.63059

C xa = C x 0 + AC yaline2 = 7.3040×10−3+ 0.0804(0.63059)2= 0.03927

K=C ya

C xa =

0 .630590 .03927

= 16.057

Determination of Aerodynamic centerM Z 0 ≈-2.8 f⃗ (1-√1−2 x⃗ f )M Z 0 ≈-2.8 x 0.021 ( 1-√1−2 x0.35 ) = -0.02532]M Z

Cyα = -0.25(1-1.6 c⃗2)M Z

Cyα = -0.25(1-1.6x 0.122) = -0.2442M Z = [cya(M Z

Cyα) + M Z 0]M Z = 0.630 (-0.24)-0.0253 = -0.17929

[94]

CHARECTERISTICS OF VERTICAL STABILIZER WITH OUTER PANEL

PARAMETERS ABBREVATION DIMENSION UNITspan of vertical stabilizer LVS 3.50 M

cross area SVS 9.01 M 2

root chord b0 VS 2.57 M

tip chord bKV S 2.57 M

mean aerodynamic chord b AVS 2.57 M

swept angle xO .25 32 DEG

aspect ratio λ 1.36 -

tapper ratio η 1 -thickness ratio CVS 0.10 -

Re = v .ba

v

Re= 835×2.57

3.46 x1.46 x10−5 = 408.28576 × 105

A =1+∂

π . λVS

A =1+0.03

3.14 .1.36 = 0.24119

C x 0friction = 2C f ηe ηm

C f = 0.087¿¿¿ = 0.087

¿¿¿ = 2.4078 ×10−3

ηm= 1+5¿¿ = 1+5¿¿ = 1.0622ηe= 1+2.τ vs+9(τ ¿¿2= 1+2.(0.10)+9(0.10¿¿2 = 1.29

C x 0friction = 2C f ηe ηm =2×2.4078 ×10−3× 1.0622×1.29=6.5985×10−3

[95]

DETERMINATION OF AERODYNAMIC CHARECTERISTICS OF FUSELAGE

Lift coefficient :C ya

α Fuselage = C yaα nose+C ya

α cylinder+C yaα rear

Lift coefficient for nose:C ya

α nose = 2(1-ηnose2 ) =2(1-0)=2

Lift coefficient for cylindrical part:C ya

α cylinder = 0Lift coefficient for rear part:

C yaα rear= -0.4(1-ηrear

2 ) = -0.4 (1-0)= -0.4

Total Lift coefficient of fuselage :C ya

α Fuselage = C yaα nose+C ya

α cylinder+C yaα rear

C yaα Fuselage = 2+0+(-0.4)=1.6

Zero Lift Angle:

α 0 fuselage=1.25 [βnose λnose

λ fuselage

+0.1βrear

λrear

λ fuselage

]

w⃗=wnose

τnose Smf

Smf =π Df 2

4 =

3.14(2.5)2

4= 4.9062 ,βrear = 4° , βnose=¿3° ¿

α 0 fuselage =1.25 [βnose λnose

λ fuselage

+0.1βrear

λrear

λ fuselage

] = 0.56°

wnose =1

15 (8+ηnose+3 ηnose

2 )lnose.Smf

wnose =1

15 (8+0+0)3.75 . 4.9062 = 9.8124

m zCya =x⃗ fnose = 1-w⃗nose

[96]

w⃗nose=9.8124

3.75× 4.9062 =0.5333

x⃗ fnose = 1-0.5333 = 0.4666For cylindrical part:C yrear =0For rear part:x f rear = Lf -0.5Lrear = 27.495Coordinate of AD centre of fuselage:

x f =−mz

α

c yaα =

C yaα nose x fnose+C ya

α cylinder x f+C yaα rear x frear

C yaα nose+C ya

α cylinder+C yaα rear

= -6.2905

Drag coefficient of fuselage bodyC xp = C xp. br + ΣC xp

C xp = 0.10984+0.16307 = 0.272911

C xp. br = C f . nλ .nm(F l . s

Sm. f)

C xp. br = 44.8007×0.95661×0.002348×1.09162 = 0.10984

C f = 0.087¿¿

C f = 0.087

37.0455 = 0.0023484

nλ = 1+1λ f

+ 1.5

λ f2

nλ = 1+1

12.25 +

1.5

(12.25)2 = 1.09126

nm = 1

√1+0.2 M ∞2

nm = 1

√1+0.2¿¿¿ = 0.95661

(F l . s

Sm. f) ≈ 4λ f [1-0.2

λnose

λ f

−0.3λrear

λ f]

(F l . s

Sm. f) ≈ 4.12.25 [1-0.2

1.512.25

−0.32.5

12.25] = 44.8007

C xp = 0.029

√C xp.b . n

׿

C xp = 0.029

√0.10984× ¿ = 0.0190645

C xp = a

√C xp.b . n

tan32( βrear

° ) Defletion of rear (β°)

[97]

C xp = 0.04

√0.10984tan

32 (4) = 2.2206 ×10−3

C xp = 0.038√ λ f Drag coefficient of cockpitC xp = 0.038√12.25 = 0.133C xp = 0.08C xp. br Drag coefficient of landing gearC xp = 0.08× 0.10984 = 8.7872 ×10−3

PROCEDURE INVOLVED IN AERODYNAMICS CALCULATION USING THE SOFTWARE

The calculation aerodynamic centre, wave drag, profile drag, lift coefficient and drag coefficient of the fuselage, wing, engines, horizontal stabilizer and vertical stabilizer are done through the software by entering the required parameters under each section. This is simplified by referring our three view diagram for certain parameters like,

distance between the wing to nose of the fuselage, distance between horizontal stabilizer to wing and the offset angle between wing and fuselage,etc.,

are found easily. Hence before proceeding to enter the values all dimensions in three views are mentioned clearly. From the results of the software graphs for various parameters of aircraft is drawn and compared.The software has seven section the details involved in calculation is as follows

MAIN PARAMETERS

[98]

FUSELAGE PARAMETERS

ENGINE PARAMETERS

[99]

WING

HORIZONTAL STABILIZER

VERTICAL STABILIZER

[100]

SECONDARY SECTIONIn The secondary section the additional drag is to 0.10

Finally we obtain the general view diagram from the software for the given data’s and it is checked with our general view diagram.

[101]

After obtaining the three view diagram the aerodynamic characteristics are obtained following the longitudinal moment of the designed aircraft. From the centre of gravity calculation we know that the cg is 18.1 so in calculation longitudinal moment this is entered and checked with values obtained.

[102]

RESULTS OF THE SOFTWARECalculation of aircraft aerodynamic characteristics==================================================================================Student: 16E02- kirubagaran Mazhalai Priyan

Aircraft type: subsonic non-manoeuvre. Configuration of aircraft: normal. Area Sch= 63.870

GEOMETRICAL PARAMETERS AND AERODYNAMIC CHARACTERISTICS OF THE ISOLATED FUSELAGE==================================================================================Lf=30.620 Df= 2.500 Lmf=12.248 Smf= 4.909 Srlf=0.0769 Fs/Sm=44.292 M*=0.934Ln= 3.750 Dn= 0.000 Lmn= 1.500 Etn= 0.000 Betn= 3.000 Shape: ellipsoidalLr= 6.250 Dr= 0.000 Lmr= 2.500 Etr= 0.000 Betr= 5.000 Shape: curvilinearAir intake - absent Sai= 0.000 Scb= 0.000 Srlcb=0.0000Cockpit Lcpt= 0.000 Scpt= 0.000Presence of the landing gear fairing----------------------------------------------------------------------------------Profile drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Point Xt 0.14019 0.13142 0.12710 0.12463 0.12311 0.12213 0.12148 0.12050H= 0.0 0.09032 0.08654 0.08405 0.08218 0.08063 0.07926 0.07800 0.07334H= 4.0 0.09464 0.09065 0.08801 0.08602 0.08437 0.08292 0.08158 0.07663H= 8.0 0.10002 0.09582 0.09302 0.09089 0.08913 0.08756 0.08612 0.08083H=11.0 0.10498 0.10063 0.09770 0.09546 0.09360 0.09194 0.09041 0.08480----------------------------------------------------------------------------------Additional profile drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.04255 0.04278 0.04295 0.04308 0.04321 0.04332 0.04343 0.08638H= 4.0 0.04233 0.04253 0.04269 0.04281 0.04292 0.04303 0.04313 0.08528H= 8.0 0.04211 0.04228 0.04241 0.04252 0.04262 0.04271 0.04281 0.08398H=11.0 0.04195 0.04209 0.04220 0.04230 0.04238 0.04246 0.04255 0.08285----------------------------------------------------------------------------------Wave drag of the isolated fuselage and its parts M*=0.934 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 0.25860Rear part 0.12471Fuselage 0.38332----------------------------------------------------------------------------------Additional wave drag of the isolated fuselage M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00000----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated fuselage and its parts Alfa0= 0.59 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.1632Cylindric 0.2005Rear part -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000 -0.4000Fuselage 1.6000 1.6000 1.6000 1.6000 1.6000 1.6000 1.6000 1.9637----------------------------------------------------------------------------------Location aerodynamic center Xac of the isolated fuselage and its parts from nose M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Nose part 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000Cylindric 0.1840Rear part 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979 0.8979Fuselage -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.1641

[103]

===================================================

GEOMETRICAL PARAMETERS AND AERODYNAMIC CHARACTERISTICS OF THE ISOLATED WING================================================================================== B0= 4.500 Bs= 4.500 Bt= 1.100 Ba= 2.800 L= 22.730 Lp=11.365 Fiwt= 0.00 Bmac= 3.144 Xmac= 2.245 (w/o Extent.)Sp= 63.644 Srl=0.9965 Lm= 8.118 Et= 4.091 (w/o Extent.)Xi00= 26.3 Xi05= 19.1 Xi10= 11.1 Xic= 22.1 Xi25= 22.8 (w/o Extent.)----------------------------------------------------------------------------------Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.150 Ñt= 0.080 Ñ= 0.136 Xc= 0.300 f= 0.020 Xf= 0.418 M*=0.776----------------------------------------------------------------------------------XB0=15.074 XBs=15.074 Fi= 2.000 distance from fuselage nose and setting angleX14=16.199 D14= 0.000 X12=17.324 D14= 0.000----------------------------------------------------------------------------------Position - wing + circular fuselage Kint= 0.644 Df= 2.500 H=-1.090 Sig=0.110Tip elements - absent----------------------------------------------------------------------------------Deceleration flow coefficient before a wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000----------------------------------------------------------------------------------Profile drag of the isolated wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.29436 0.27113 0.25743 0.24863 0.24270 0.23857 0.23567 0.23071H= 0.0 0.00577 0.00549 0.00528 0.00511 0.00497 0.00484 0.00472 0.00428H= 4.0 0.00604 0.00575 0.00554 0.00537 0.00522 0.00508 0.00496 0.00450H= 8.0 0.00636 0.00606 0.00585 0.00568 0.00552 0.00538 0.00525 0.00477H=11.0 0.00665 0.00635 0.00613 0.00595 0.00580 0.00565 0.00552 0.00502----------------------------------------------------------------------------------Wave drag of the isolated wing M*=0.776 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.17041----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated wing Alfa0= -1.78 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 4.0314 4.1076 4.2227 4.3875 4.6210 4.9590 5.4772 5.3445----------------------------------------------------------------------------------Location aerodynamic center Xac isolated wing from nose of side chord M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.6637 0.6637 0.6637 0.6637 0.6637 0.6637 0.6637 0.8478==================================================================================

[104]

GEOMETRICAL PARAMETERS AND A/D CHARACTERISTICS OF THE ISOLATED HORIZONTAL TAIL================================================================================== B0= 2.480 Bs= 2.480 Bt= 1.240 Ba= 1.860 L= 7.550 Lp= 3.775 Bmac= 1.929 Xmac= 0.748 (w/o Extent.)Sp= 14.043 Srl=0.2199 Lm= 4.059 Et= 2.000 (w/o Extent.)Xi00= 24.0 Xi05= 15.7 Xi10= 6.7 Xic= 19.1 Xi25= 20.0 (w/o Extent.)

Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.120 Ñt= 0.120 Ñ= 0.120 Xc= 0.300 f= 0.020 Xf= 0.417 M*=0.786----------------------------------------------------------------------------------XB0=28.900 XBs=28.900 Fi= 0.000 distance from fuselage nose and setting angleX14=29.520 D14= 0.000 X12=30.140 D14= 0.000 X1= 9.797 B1= 3.935 Xht=13.321 Yht= 4.500 S*/Sw= 0.808----------------------------------------------------------------------------------Position - Tee-tail horizontal tail Sig=0.000 Kint= 0.000Tip elements - absent----------------------------------------------------------------------------------Deceleration flow coefficient before a horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842----------------------------------------------------------------------------------Profile drag of the isolated horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.33896 0.31160 0.29423 0.28243 0.27406 0.26794 0.26339 0.00000H= 0.0 0.00583 0.00555 0.00534 0.00516 0.00501 0.00487 0.00474 0.00603H= 4.0 0.00609 0.00580 0.00559 0.00541 0.00525 0.00511 0.00498 0.00635H= 8.0 0.00641 0.00610 0.00589 0.00571 0.00555 0.00540 0.00526 0.00677H=11.0 0.00672 0.00638 0.00616 0.00597 0.00581 0.00566 0.00552 0.00718----------------------------------------------------------------------------------Wave drag of the isolated horizontal tail M*=0.786 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.09897----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated horizontal tail Alfa0= -1.77 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 3.2819 3.3328 3.4090 3.5170 3.6678 3.8818 4.2007 4.3515----------------------------------------------------------------------------------Location aerodynamic center Xac isolated horizontal tail from nose of side chord M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.4472 0.4472 0.4472 0.4473 0.4474 0.4478 0.4489 0.6794==================================================================================

[105]

GEOMETRICAL PARAMETERS AND A/D CHARACTERISTICS OF THE ISOLATED VERTICAL TAIL==================================================================================Vertical tail - central (1pcs.) B0= 2.570 Bs= 2.570 Bt= 2.570 Ba= 2.570 L= 3.500 Lp= 3.500 Bmac= 2.570 Xmac= 1.085 (w/o Extent.)Sp= 8.995 Srl=0.1408 Lm= 1.362 Et= 1.000 (w/o Extent.)Xi00= 31.8 Xi05= 31.8 Xi10= 31.8 Xic= 31.8 Xi25= 31.8 (w/o Extent.)----------------------------------------------------------------------------------Type of airfoil - classical Kp= 2.1 m= 0.350 Ñs= 0.100 Ñt= 0.100 Ñ= 0.100 Xc= 0.300 f= 0.000 Xf= 0.000 M*=0.837----------------------------------------------------------------------------------XB0=26.730 XBs=26.730 Fi= 0.000 distance from fuselage nose and setting angleX14=27.372 D14= 0.000 X12=28.015 D14= 0.000----------------------------------------------------------------------------------Position a= 1.250 b= 0.000 Lm= 1.362----------------------------------------------------------------------------------

Deceleration flow coefficient before a vertical tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H= 4.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H= 8.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809H=11.0 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99809----------------------------------------------------------------------------------Profile drag of the isolated vertical tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.24813 0.22775 0.21552 0.20756 0.20211 0.19827 0.19551 0.00000H= 0.0 0.00566 0.00534 0.00511 0.00492 0.00476 0.00462 0.00449 0.00502H= 4.0 0.00595 0.00561 0.00537 0.00518 0.00501 0.00486 0.00472 0.00529H= 8.0 0.00631 0.00595 0.00570 0.00550 0.00532 0.00517 0.00502 0.00562H=11.0 0.00665 0.00627 0.00601 0.00580 0.00561 0.00545 0.00529 0.00595----------------------------------------------------------------------------------

Wave drag of the isolated vertical tail M*=0.837 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.02900

GEOMETRICAL PARAMETERS AND AERODYNAMIC CHARACTERISTICS OF ISOLATED ENGINE NACELLE==================================================================================

Engine nacelles in a tail part of a fuselage Nen= 2 Kint= 1.600Len= 4.018 Den= 1.031 Dcb= 1.031 Lm= 3.896 S= 0.835 Srl=0.0131 Fs/Sm=15.583Distance from fuselage nose Lfen=22.182 M*= 0.743Type of engine - Turbo-jet engine----------------------------------------------------------------------------------Profile drag of the isolated engine nacelle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xt 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 0.0 0.04601 0.04496 0.04461 0.04455 0.04463 0.04477 0.04492 0.04530H= 4.0 0.04867 0.04748 0.04705 0.04695 0.04700 0.04712 0.04725 0.04759H= 8.0 0.05214 0.05076 0.05023 0.05007 0.05008 0.05017 0.05028 0.05055H=11.0 0.05554 0.05395 0.05331 0.05309 0.05306 0.05312 0.05321 0.05340----------------------------------------------------------------------------------

[106]

Additional profile drag of the isolated engine nacelle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Wave drag of the isolated engine nacelle M*=0.743 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00861----------------------------------------------------------------------------------Derivative dCy/dAl of the isolated engine nacelle dCy/dAl= 0.0000==================================================================================Critical Mach number of aircraft M*=0.95 min{0.776,0.786,0.837,0.934,0.743}= 0.738

DRAG OF THE AIRCRAFT PATRS AND ADDITIONAL ELEMENTS INTO AIRCRAFT SYSTEM==================================================================================Profile drag of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.01021 0.00994 0.00976 0.00963 0.00952 0.00942 0.00933 0.01228H= 4.0 0.01053 0.01024 0.01004 0.00990 0.00978 0.00968 0.00958 0.01244H= 8.0 0.01092 0.01061 0.01041 0.01025 0.01013 0.01001 0.00991 0.01267H=11.0 0.01129 0.01097 0.01075 0.01059 0.01045 0.01033 0.01022 0.01289----------------------------------------------------------------------------------Wave drag of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.02946----------------------------------------------------------------------------------Profile drag of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00575 0.00547 0.00526 0.00509 0.00495 0.00482 0.00470 0.00426H= 4.0 0.00602 0.00573 0.00552 0.00535 0.00520 0.00506 0.00494 0.00448H= 8.0 0.00633 0.00604 0.00583 0.00566 0.00550 0.00536 0.00523 0.00476H=11.0 0.00663 0.00632 0.00611 0.00593 0.00578 0.00563 0.00550 0.00501----------------------------------------------------------------------------------Interference profile drag of (wing + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------Wave drag of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.16980H= 4.0 0.16980H= 8.0 0.16980H=11.0 0.16980----------------------------------------------------------------------------------

Interference wave drag of (wing + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000H= 4.0 0.00000 0.00000H= 8.0 0.00000 0.00000H=11.0 0.00000 0.00000----------------------------------------------------------------------------------

[107]

Profile drag of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00128 0.00122 0.00117 0.00113 0.00110 0.00107 0.00104 0.00124H= 4.0 0.00134 0.00128 0.00123 0.00119 0.00116 0.00112 0.00109 0.00131H= 8.0 0.00141 0.00134 0.00129 0.00125 0.00122 0.00119 0.00116 0.00140H=11.0 0.00148 0.00140 0.00135 0.00131 0.00128 0.00124 0.00121 0.00148----------------------------------------------------------------------------------Interference profile drag of (horizontal tail + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 4.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H= 8.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000H=11.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------

Wave drag of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.02042H= 4.0 0.02042H= 8.0 0.02042H=11.0 0.02042----------------------------------------------------------------------------------Interference wave drag of (horizontal tail + fuselage) system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00000 0.00000H= 4.0 0.00000 0.00000H= 8.0 0.00000 0.00000H=11.0 0.00000 0.00000----------------------------------------------------------------------------------Profile drag of the vertical tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00080 0.00075 0.00072 0.00069 0.00067 0.00065 0.00063 0.00071H= 4.0 0.00084 0.00079 0.00076 0.00073 0.00071 0.00068 0.00066 0.00074H= 8.0 0.00089 0.00084 0.00080 0.00077 0.00075 0.00073 0.00071 0.00079H=11.0 0.00094 0.00088 0.00085 0.00082 0.00079 0.00077 0.00075 0.00084----------------------------------------------------------------------------------Wave drag of the vertical tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00408H= 4.0 0.00408H= 8.0 0.00408H=11.0 0.00408----------------------------------------------------------------------------------

Profile drag of the engine nacelle into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00193 0.00188 0.00187 0.00186 0.00187 0.00187 0.00188 0.00190H= 4.0 0.00204 0.00199 0.00197 0.00196 0.00197 0.00197 0.00198 0.00199H= 8.0 0.00218 0.00212 0.00210 0.00210 0.00210 0.00210 0.00210 0.00212H=11.0 0.00232 0.00226 0.00223 0.00222 0.00222 0.00222 0.00223 0.00223----------------------------------------------------------------------------------Wave drag of the engine nacelle into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.00023----------------------------------------------------------------------------------

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Profile drag of the aircraft (with out additional elements) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.01997 0.01926 0.01878 0.01841 0.01811 0.01784 0.01759 0.02039H= 4.0 0.02076 0.02002 0.01952 0.01913 0.01881 0.01852 0.01826 0.02097H= 8.0 0.02174 0.02096 0.02044 0.02003 0.01969 0.01939 0.01911 0.02173H=11.0 0.02266 0.02184 0.02129 0.02087 0.02052 0.02020 0.01990 0.02244----------------------------------------------------------------------------------Wave drag of the aircraft (with out additional elements) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.22398H= 4.0 0.22398H= 8.0 0.22398H=11.0 0.22398----------------------------------------------------------------------------------Summarized drag of the aircraft's additional elemetns KdCx= 0.100 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.00200 0.00193 0.00188 0.00184 0.00181 0.00178 0.00176 0.02444H= 4.0 0.00208 0.00200 0.00195 0.00191 0.00188 0.00185 0.00183 0.02450H= 8.0 0.00217 0.00210 0.00204 0.00200 0.00197 0.00194 0.00191 0.02457H=11.0 0.00227 0.00218 0.00213 0.00209 0.00205 0.00202 0.00199 0.02464----------------------------------------------------------------------------------Drag coefficient of aircraft at zero lift Cya=0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881H= 4.0 0.02283 0.02202 0.02147 0.02105 0.02069 0.02038 0.02008 0.26945H= 8.0 0.02391 0.02306 0.02248 0.02204 0.02166 0.02133 0.02102 0.27028H=11.0 0.02492 0.02402 0.02342 0.02296 0.02257 0.02222 0.02189 0.27107----------------------------------------------------------------------------------Wave drag coefficient of aircraft from M*=0.738 to M=1.2 M=0.5 M=0.6 M=0.7 M=0.8 M=0.9 M=1.0 M=1.1 M=1.2H= 0.0 0.00198 0.02860 0.09406 0.17779 0.22398H= 4.0 0.00198 0.02860 0.09406 0.17779 0.22398H= 8.0 0.00198 0.02860 0.09406 0.17779 0.22398H=11.0 0.00198 0.02860 0.09406 0.17779 0.22398----------------------------------------------------------------------------------Drag coefficient of aircraft from M*=0.738 to M=1.2 M=0.5 M=0.6 M=0.7 M=0.8 M=0.9 M=1.0 M=1.1 M=1.2H= 0.0 0.02664 0.05831 0.12880 0.21758 0.26881H= 4.0 0.02738 0.05902 0.12949 0.21824 0.26945H= 8.0 0.02831 0.05992 0.13037 0.21910 0.27028H=11.0 0.02918 0.06077 0.13120 0.21991 0.27107

==================================================================================

LIFT OF THE AIRCRAFT PATRS INTO AIRCRAFT SYSTEM==================================================================================Average value of deceleration flow coefficient before a wing M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000----------------------------------------------------------------------------------Interference coefficient: wing + fuselage kAl0=1.04423 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 kAl 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423 1.04423DkAl 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 0.04619 kFi 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkFi 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423 0.04423----------------------------------------------------------------------------------

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Average value of deceleration flow coefficient before a horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.93842----------------------------------------------------------------------------------Interference coefficient: horizontal tail + fuselage kAl0=1.00000 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 kAl 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkAl 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 kFi 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000DkFi 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000----------------------------------------------------------------------------------

Downwash before horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20eps0 0.497° 0.504° 0.515° 0.531° 0.554° 0.587° 0.640° 0.546°epsAl 0.1357 0.1377 0.1407 0.1451 0.1513 0.1605 0.1748 0.1492----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 4.38036 4.46318 4.58822 4.76728 5.02104 5.38826 5.95129 5.80711Alfa0 -3.62° -3.62° -3.62° -3.62° -3.62° -3.62° -3.62° -3.62°----------------------------------------------------------------------------------

Derivative dCy/dAl & Alfa0 of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 0.62366 0.63187 0.64406 0.66109 0.68441 0.71651 0.76219 0.76384Alfa0 -1.47° -1.47° -1.46° -1.45° -1.43° -1.41° -1.37° -1.44°----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 0.12297 0.12297 0.12297 0.12297 0.12297 0.12297 0.12297 0.15092Alfa0 0.59° 0.59° 0.59° 0.59° 0.59° 0.59° 0.59° 0.59°----------------------------------------------------------------------------------Derivative dCy/dAl engine nacelle into aircraft system dCy/dAl= 0.00000----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the aircraft with out horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 4.50333 4.58615 4.71119 4.89025 5.14401 5.51123 6.07425 5.95803Alfa0 -3.50° -3.50° -3.51° -3.51° -3.51° -3.52° -3.53° -3.51°----------------------------------------------------------------------------------Derivative dCy/dAl & Alfa0 of the aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dCy/dAl 5.12699 5.21802 5.35525 5.55134 5.82842 6.22773 6.83644 6.72188Alfa0 -3.12° -3.12° -3.12° -3.13° -3.14° -3.15° -3.17° -3.15°----------------------------------------------------------------------------------

Lift coefficient of the aircraft Cya=dCy/dAl(Al-Al0) (line section) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Al= 0.0° 0.27875 0.28399 0.29190 0.30323 0.31927 0.34247 0.37800 0.36911Al=10.0° 1.17358 1.19471 1.22657 1.27212 1.33652 1.42941 1.57118 1.54229Al=20.0° 2.06841 2.10542 2.16124 2.24101 2.35377 2.51636 2.76437 2.71548----------------------------------------------------------------------------------Maximum lift coefficient of aircraft and stalling angle M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya max 1.23511 1.19831 1.15952 1.11875 1.07599 1.03126 0.98453 Alfa st 12.19° 11.54° 10.78° 9.92° 8.94° 7.84° 6.58° ----------------------------------------------------------------------------------Polar coefficient A M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.04750 0.04748 0.04746 0.04744 0.04740 0.04734 0.04726 0.14879----------------------------------------------------------------------------------

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Maximum lift to drag ratio Kmax M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 15.480 15.765 15.967 16.131 16.273 16.406 16.537 2.500H= 4.0 15.184 15.463 15.663 15.824 15.966 16.098 16.229 2.497H= 8.0 14.837 15.111 15.306 15.464 15.604 15.734 15.863 2.493H=11.0 14.532 14.805 14.996 15.151 15.288 15.416 15.544 2.490----------------------------------------------------------------------------------

Most efficiently lift coefficient Cya m.e. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20H= 0.0 0.68004 0.66794 0.65974 0.65345 0.64826 0.64376 0.63977 1.34412H= 4.0 0.69331 0.68097 0.67257 0.66610 0.66073 0.65606 0.65191 1.34573H= 8.0 0.70951 0.69685 0.68825 0.68161 0.67608 0.67124 0.66693 1.34780H=11.0 0.72440 0.71126 0.70246 0.69568 0.69003 0.68508 0.68066 1.34976----------------------------------------------------------------------------------

Polar of aircraft Cxa = Cxo + A·Cya^2 at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881Cya= 0.2 0.02386 0.02308 0.02256 0.02215 0.02181 0.02151 0.02123 0.27476Cya= 0.4 0.02956 0.02878 0.02825 0.02784 0.02750 0.02719 0.02691 0.29261Cya= 0.6 0.03906 0.03828 0.03775 0.03733 0.03698 0.03666 0.03636 0.32237Cya= 0.8 0.05236 0.05157 0.05104 0.05061 0.05025 0.04992 0.04959 0.36403Cya= 1.0 0.06946 0.06867 0.06812 0.06769 0.06731 0.06696 0.41760Cya= 1.2 0.09036 0.48306----------------------------------------------------------------------------------Additional induced drag of aircraft dCxi at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 Cya= 0.2 0.00004 0.00005 0.00005 0.00006 0.00006 0.00007 0.00008 Cya= 0.4 0.00034 0.00038 0.00042 0.00047 0.00052 0.00060 0.00425 Cya= 0.6 0.00120 0.00132 0.00146 0.00164 0.00186 0.00318 0.02538 Cya= 0.8 0.00305 0.00338 0.00379 0.00430 0.00583 0.03290 0.06093 Cya= 1.0 0.00689 0.00782 0.00905 0.01537 0.04875 0.08417 Cya= 1.2 0.01854 Cya= Max 1.43894 1.29362 1.13961 0.97436 0.79364 0.58919 0.33912 ----------------------------------------------------------------------------------

Polar of aircraft Cxa = Cxo + A·Cya^2 + dCxi at H= 0.0 M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.02196 0.02118 0.02066 0.02026 0.01992 0.01962 0.01934 0.26881Cya= 0.2 0.02391 0.02313 0.02261 0.02221 0.02188 0.02159 0.02132 0.27476Cya= 0.4 0.02991 0.02916 0.02867 0.02831 0.02803 0.02779 0.03116 0.29261Cya= 0.6 0.04026 0.03960 0.03921 0.03897 0.03884 0.03984 0.06174 0.32237Cya= 0.8 0.05541 0.05495 0.05483 0.05492 0.05608 0.08282 0.11052 0.36403Cya= 1.0 0.07636 0.07649 0.07717 0.08306 0.11606 0.15113 0.41760Cya= 1.2 0.10890 0.48306Cya= Max 0.13442 0.12937 0.12878 0.14872 0.16692 0.18317 0.19810 ----------------------------------------------------------------------------------K max 14.924 15.159 15.303 15.396 15.455 15.486 13.460 2.500Ñya m.e. 0.63109 0.61672 0.60513 0.59448 0.58391 0.57286 0.33914 1.34413Alf m.e. 3.94° 3.65° 3.35° 3.01° 2.60° 2.12° -0.33° 8.31°==================================================================================

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LONGITUDINAL MOMENT AND AERODYNAMIC CENTER LOCATION OF AIRCRAFT PATS INTO AIRCRAFT SYSTEM==================================================================================Derivative dMz/dAl & aerodynamic center Xac/Lf of the wing into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -2.5837 -2.6325 -2.7063 -2.8119 -2.9616 -3.1782 -3.5103 -3.5823Xac/Lf 0.5898 0.5898 0.5898 0.5898 0.5898 0.5898 0.5898 0.6169----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of the horizontal tail into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -0.6112 -0.6193 -0.6312 -0.6479 -0.6708 -0.7022 -0.7471 -0.7630Xac/Lf 0.9800 0.9800 0.9800 0.9801 0.9801 0.9801 0.9802 0.9989----------------------------------------------------------------------------------

Derivative dMz/dAl & aerodynamic center Xac/Lf of the fuselage into aircraft system M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl 0.0276 0.0276 0.0276 0.0276 0.0276 0.0276 0.0276 0.0248Xac/Lf -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.2245 -0.1641----------------------------------------------------------------------------------Derivative dMz/dAl engine nacelle into aircraft system dMz/dAl= 0.0000Position of the engine nacelle from fuselage nose Xìãä/Lf= 0.7244----------------------------------------------------------------------------------Derivative dMz/dAl & aerodynamic center Xac/Lf of aircraft w/o horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -2.5561 -2.6049 -2.6787 -2.7843 -2.9340 -3.1506 -3.4827 -3.5576Xac/Lf 0.5676 0.5680 0.5686 0.5694 0.5704 0.5717 0.5733 0.5971----------------------------------------------------------------------------------

Derivative dMz/dAl & aerodynamic center Xac/Lf of aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dMz/dAl -3.1673 -3.2242 -3.3099 -3.4322 -3.6047 -3.8528 -4.2298 -4.3205Xac/Lf 0.6178 0.6179 0.6181 0.6183 0.6185 0.6187 0.6187 0.6428----------------------------------------------------------------------------------Aerodynamic center Xac/Bmac of aircraft w/o horizontal tail relative to MAC L.E. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xac/Bmac 0.0194 0.0233 0.0289 0.0365 0.0463 0.0590 0.0754 0.3067----------------------------------------------------------------------------------Aerodynamic center Xac/Bmac of aircraft with horizontal tail relative to MAC L.E. M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Xac/Bmac 0.5080 0.5092 0.5109 0.5128 0.5149 0.5166 0.5171 0.7513----------------------------------------------------------------------------------

Shift of the aerodynamic center of aircraft dXac/Bmac M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20dXac/Bmac 0.4886 0.4859 0.4819 0.4763 0.4685 0.4576 0.4417 0.4446----------------------------------------------------------------------------------Longitudinal moment mz0 of wing and (fuselage+wing) system (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0iz.w -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333 -0.0333mz0f(w) 0.0932 0.0933 0.0933 0.0934 0.0935 0.0937 0.0939 0.0934mz0* 0.0366 0.0366 0.0366 0.0367 0.0368 0.0369 0.0371 0.0367----------------------------------------------------------------------------------

Longitudinal moment mz0 of the parts of the aircraft w/o horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0w’ -0.0599 -0.0599 -0.0599 -0.0600 -0.0601 -0.0602 -0.0603 -0.0706mz0f’ -0.0097 -0.0097 -0.0097 -0.0097 -0.0097 -0.0098 -0.0098 -0.0115----------------------------------------------------------------------------------

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Longitudinal moment mz0 of the aircraft w/o horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454----------------------------------------------------------------------------------Longitudinal moment mz0 of the parts of the aircraft with horizontal tail M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20mz0w 0.0752 0.0770 0.0796 0.0833 0.0884 0.0954 0.1056 0.0923mz0h.t. -0.0566 -0.0571 -0.0579 -0.0589 -0.0603 -0.0620 -0.0641 -0.0665mz0f -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0103 -0.0121----------------------------------------------------------------------------------Longitudinal moment mz0 of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504================================================================================== © Holjavko V.I. Aerodynamic characteristics of aircraft, 1991 - 1998 © Chmovzh V.V. 1991 - 2011 Âåðñèÿ 4.11b îò 01.11.2011

LONGITUDINAL MOMEMT OF THE AIRCRAFT RELATIVE TO MASS GRAVITY XMG=18.100 XMG/LF=0.591==================================================================================Longitudinal momemt of the aircraft with out horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454Cya= 0.2 0.0128 0.0120 0.0109 0.0094 0.0074 0.0049 0.0016 -0.0570Cya= 0.4 0.0586 0.0571 0.0548 0.0518 0.0478 0.0428 0.0362 -0.0687Cya= 0.6 0.1044 0.1021 0.0987 0.0942 0.0883 0.0807 0.0708 -0.0803Cya= 0.8 0.1502 0.1471 0.1426 0.1365 0.1287 0.1186 0.1054 -0.0920Cya= 1.0 0.1961 0.1921 0.1865 0.1789 0.1691 0.1565 -0.1036Cya= 1.2 0.2419 -0.1153

Longitudinal momemt of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504Cya= 0.2 -0.0071 -0.0060 -0.0044 -0.0021 0.0013 0.0064 0.0144 -0.0502Cya= 0.4 -0.0590 -0.0582 -0.0569 -0.0550 -0.0520 -0.0473 -0.0393 -0.1508Cya= 0.6 -0.1109 -0.1103 -0.1094 -0.1079 -0.1053 -0.1009 -0.0930 -0.2514Cya= 0.8 -0.1628 -0.1625 -0.1619 -0.1608 -0.1586 -0.1545 -0.1468 -0.3520Cya= 1.0 -0.2147 -0.2147 -0.2144 -0.2136 -0.2119 -0.2082 -0.4525Cya= 1.2 -0.2666 -0.5531==================================================================================

Longitudinal momemt of the aircraft relative to mass gravity Xmg=18.100 Xmg/Lf=0.591==================================================================================Longitudinal momemt of the aircraft with out horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0330 -0.0454Cya= 0.2 0.0128 0.0120 0.0109 0.0094 0.0074 0.0049 0.0016 -0.0570Cya= 0.4 0.0586 0.0571 0.0548 0.0518 0.0478 0.0428 0.0362 -0.0687Cya= 0.6 0.1044 0.1021 0.0987 0.0942 0.0883 0.0807 0.0708 -0.0803Cya= 0.8 0.1502 0.1471 0.1426 0.1365 0.1287 0.1186 0.1054 -0.0920Cya= 1.0 0.1961 0.1921 0.1865 0.1789 0.1691 0.1565 -0.1036Cya= 1.2 0.2419 -0.1153----------------------------------------------------------------------------------

[113]

Longitudinal momemt of the aircraft with horizontal tail (relative to MAC) M=0.20 M=0.30 M=0.40 M=0.50 M=0.60 M=0.70 M=0.80 M=1.20Cya= 0.0 0.0449 0.0461 0.0480 0.0508 0.0546 0.0600 0.0682 0.0504Cya= 0.2 -0.0071 -0.0060 -0.0044 -0.0021 0.0013 0.0064 0.0144 -0.0502Cya= 0.4 -0.0590 -0.0582 -0.0569 -0.0550 -0.0520 -0.0473 -0.0393 -0.1508Cya= 0.6 -0.1109 -0.1103 -0.1094 -0.1079 -0.1053 -0.1009 -0.0930 -0.2514Cya= 0.8 -0.1628 -0.1625 -0.1619 -0.1608 -0.1586 -0.1545 -0.1468 -0.3520Cya= 1.0 -0.2147 -0.2147 -0.2144 -0.2136 -0.2119 -0.2082 -0.4525Cya= 1.2 -0.2666 -0.5531==================================================================================

The results obtained from this calculation are further used to draw graphs of aircraft polar and other aerodynamic characteristics.

AIRCRAFT DESIGN PARAMETERS ON AERODYNAMIC CHARACTERISTICS

dCy/dAl Cya = 4.86167

Alfa0 = -3.16

Cya max = 1.23511

A = 0.04772

Cxa = 0.02196

Calculation of lift coefficient for the whole airplane

Cyaα= Cyafα .Sf+¿ Cyawα .Sw+¿ CyaHSα .SHS

Cyaα = 4.86167 ( from software )

α o = 1

C yaα.(α of. Cyafα .Sf + α ow. Cyawα .Sw) + (α oHS.CyaHSα .SHS)

α o ¿−¿3.16 (from software )

Using equation, Cya = Cyaα .(α−αo)×π

180 we get,

S.No α Cyaα α0 Cya

1 0

4.8616

7 -3.16 0.26799

2 2

4.8616

7 -3.16

0.43760

4

3 4 4.8616 -3.16 0.60721

[114]

7 8

4 6

4.8616

7 -3.16

0.77683

2

5 8

4.8616

7 -3.16

0.94644

6

6 10

4.8616

7 -3.16 1.11606

7 12

4.8616

7 -3.16

1.28567

4

8 14

4.8616

7 -3.16

1.45528

8

9 16

4.8616

7 -3.16

1.62490

2

10 18

4.8616

7 -3.16

1.79451

6

11 20

4.8616

7 -3.16

1.96412

9

CALCULATION OF LIFT COEFFICIENT WHILE TAKE-OFF AND LANDING

Cya = Cyaα .(α−αo)×π

180

Cyaα = 4.86167 ( from software ) , α o= -3.16

( from software )

FOR TAKE-OFF,

∆ αo take-off = -8 (∆ αo take-off = −¿6°......−¿10°)

α o = ¿−¿3.16– 8 = -11.16

Alpha zero = -3.16 -8 = -11.16

Using equation, Cya take-off = Cyaα .(α−αo)×π

180 we get,

α Cyaα α0 Cya

-6 4.8616 -11.16 0.43760

[115]

7 4

-4

4.8616

7 -11.16

0.60721

8

-2

4.8616

7 -11.16

0.77683

2

0

4.8616

7 -11.16

0.94644

6

2

4.8616

7 -11.16 1.11606

4

4.8616

7 -11.16

1.28567

4

6

4.8616

7 -11.16

1.45528

8

8

4.8616

7 -11.16

1.62490

2

10

4.8616

7 -11.16

1.79451

6

12

4.8616

7 -11.16

1.96412

9

14

4.8616

7 -11.16

2.13374

3

16

4.8616

7 -11.16

2.30335

7

18

4.8616

7 -11.16

2.47297

1

20

4.8616

7 -11.16

2.64258

5

FOR LANDING

[116]

For landing,

∆ αo landing = −10(∆ αo landing = −¿10°......−¿15°)

Alpha zero = -3.16 -12 = -15.16

Cyaα = 4.86167 ( from software )

CALCULATION OF ZERO DRAG COEFFICIENT FOR TAKE-OFF AND

LANDING

As we know, Cxa = Cxao + A.Cya2

So, Cxa = (Cxao flight + ∆Cxao Flaps + ∆Cxao Landing gear) + A.Cya2

Cxa = 0.02196

A = 0.04772

Using the given coefficients,

α Cyaα α0 Cya

-12 4.86167 -15.16 0.26799

-10 4.86167 -15.16 0.437604

-8 4.86167 -15.16 0.607218

-6 4.86167 -15.16 0.776832

-4 4.86167 -15.16 0.946446

-2 4.86167 -15.16 1.11606

0 4.86167 -15.16 1.285674

2 4.86167 -15.16 1.455288

4 4.86167 -15.16 1.624902

6 4.86167 -15.16 1.794516

8 4.86167 -15.16 1.964129

10 4.86167 -15.16 2.133743

12 4.86167 -15.16 2.303357

14 4.86167 -15.16 2.472971

16 4.86167 -15.16 2.642585

18 4.86167 -15.16 2.812199

20 4.86167 -15.16 2.981813

[117]

FOR TAKE – OFF,

∆ C xao FlapsC xao

= 0.4.......0.5 = 0.45

∆ C xao Landing gearC xao

= 0.5.....0.6 = 0.55

Cxa = Cxao.(1+∆ C xao FlapsC xao

+ ∆C xao Landing gearC xao ) + A.Cya

2

FOR LANDING,

Cxa = 0.02196

A = 0.04772

∆ C xao FlapsC xao

= 1.0.........1.5 = 1.25

∆ C xao Landing gearC xao

= 0.5.........0.6 = 0.55

Cxa = Cxao.(1+∆ C xao FlapsC xao

+ ∆C xao Landing gearC xao ) + A.Cya

2

S.No Cya Cxa0 A Cxa

1 0

0.0219

6

0.0477

2

0.06148

8

2 0.2

0.0219

6

0.0477

2

0.06339

7

3 0.4 0.0219 0.0477 0.06912

S.No Cya Cxa0 A Cxa

1 0 0.02196 0.04772 0.04392

2 0.2 0.02196 0.04772 0.045829

3 0.4 0.02196 0.04772 0.051555

4 0.6 0.02196 0.04772 0.061099

5 0.8 0.02196 0.04772 0.074461

6 1 0.02196 0.04772 0.09164

7 1.2 0.02196 0.04772 0.112637

[118]

6 2 3

4 0.6

0.0219

6

0.0477

2

0.07866

7

5 0.8

0.0219

6

0.0477

2

0.09202

9

6 1

0.0219

6

0.0477

2

0.10920

8

7 1.2

0.0219

6

0.0477

2

0.13020

5

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.02

0.04

0.06

0.08

0.1

0.12

0.14

For take - offFor landing

-15-10 -5 0 5 10 15 20 250

0.20.40.60.8

11.21.41.61.8

22.22.42.62.8

33.2

For flightFor take-offFor landing

Conclusion: The aerodynamic characteristics of the designed aircraft are calculated both

manually and through the software.

[119]

DESIGN SECTION PART 3

3.1. INTEGRATED DESIGNING AND COMPUTER AIDED MODELING OF DESIGNED AIRCRAFT

3.2. WING LOAD CALCULATION

[120]

3.1.INTEGRATED DESIGNING AND COMPUTER AIDED MODELING WING OF DESIGNED AIRCRAFTFuselage modeling:Fuselage parameters are calculated in PART 1 AND the design Parameters of fuselage are used for the surface modeling in Unigraphics NX6 the surface model of the fuselage developed in NX 6 is shown below

Fuselage surface modelWing designing:Wing is the main lifting surface in the aircraft so it should maintain aerodynamic shape for that we need to consider the airfoil shapes for the wing the Wing airfoil are generated by using profile 2 software and I used NACA 2415 Mod 40-30 airfoil for wing root and wing tips. The airfoils are imported in to Unigraphics as shown below:

Airfoils imported from profili to NX6

[121]

The airfoil are moved from one another according to wing parameters and by using mesh surface option created the wing surface under rules mesh and also high lifting devices like flaps and control surfaces like ailerons are also represented in the design as shown in bellow figure.

Wing surface

Designing of tail section:Horizontal and vertical stabilizers are generated as same procedure used in wing designing. My configuration of tail is T tail the horizontal stabilizer is located on vertical stabilizer .tail unit control surfaces are calculated in PART-1 from those values elevator and rudder surfaces are created. The surface design of tail part is shown below.

Tail surface

[122]

Assembling the unitsAfter creating the aircraft fuselage, wing and tail unit the individual units are assembled together. The assembling process depends up on location of center of gravity of aircraft .so the center of gravity of the aircraft which is calculated already in first section is used here.As like wings tail unit is also mounted on the fuselage in same manner The corresponding surface model of my aircraft is done and represented.

Surface model of aircraft

[123]

Top view

Front view

Side view

[124]

WING LOAD CALCULATION

WING’S GENERAL DATA

WING’S GEOMETRICAL DATAbrs = 4.5mbts = 1.1mLw = 22.73m

Length of half section of wing = Lw

2 =11.365 m

Sw = 0.5 (br + bt) Lw = 0.5(4.5 + 1.1) 22.73

=63.64 m 2

Technical DataTake- off Weight Gt = 383278Kg Wing Span Lw = 22.73 mWing Area Sw = 63.64 m2

Cruising Speed VC = 835 Km/hCruising Altitude HC = 10000 mWing sweep till 0.25 of chord χ0.25 = 22.730

[125]

CALCULATION OF THE DISTRIBUTED FUEL LOAD ON A PLANE WING

Approximately it is possible to consider, that linear load from fuel qf changes linearly on length of each tank separately. For definition of this function it is necessary to find sizes qf

on borders of tanks. With this purpose it is necessary to construct cross section of a wing with spars and to find the area of cross section, which is occupied by fuel –Sf.

It is possible to make manually or with the help of the program "Compass". As a first approximation Sf. is equal:

where μ=0.9 – is factor, which takes into account, what not all cross section is engaged in fuel (part is engaged in a structure, fixture, air for overpressure of fuel), Н1, Н2 - height accordingly front and rear spars, and a- distance between spars. The linear load qf is under the formula:

[N/m],Where р=800kg/m3 – is density of fuel, g=9.81 m/с2 - is acceleration of free fall.*Location of Front Spar = 0.25 x chord of the Wing.*Location of Rear Spar = 0.74 x chord*Wing’s centre of gravity centre cross-section = 0.40 x chord from leading edge*Landing Gear base = Distance between two landing gears fitted on each side of wing = (0.20…..0.25) x Length of Wing (LW = 22.73) = 0.20 x 22.73

= 4.546 m*Distance b/w Wing Leading edge to LG centre of gravity (bl) = 0.3 x Wing chord (b) = 0.3 x 4.5 = 1.35 m*Weight of primary structure of main LG = 0.45 x Weight of LG = 0.45 x 1400 kg = 630 kg *Weight of primary structure of Nose LG = 0.45 x Weight of Nose LG

= 0.45 x 697.33 kg = 311.0915 kg.

[126]

DISTRIBUTED FUEL LOAD ON WING

Sf= Area of cross-section of wing with spars.qf = Linear Load.qf

u= Ultimate Linear Load.Gwf = Weight of fuel in tank.Xf= Position of CG for fuel in cross-section.Sf =µ.(H1+ H2/2).µ = 0.9H1 = Height of front sparsH2 = Height of Rear sparsa = Distance b/w two spars.

[Total no of fuel tanks = 2], one at 0.2 & another one at 0.5.Equivalent wing is drawn, Airfoil characteristics are plotted, from that the value of H1& H2 is found.

At 0.2, chord length = 3.82, NACA 2415H1 = 0.5680 m ; H2 = 0.3247 ma = Location of rear spar – Location of front spar.Location of Front spar = 0.25 x 3.82 = 0.955 m.Location of Rear spar = 0.74 x 3.82 = 2.826 m.*a = 2.826 – 0.955 = 1.871 m.

,*Sf fuel tank 1 at 0.2 = 0.9 x (0.3247 + 0.5680/2) x 1.871 = 0.9 x 0.44 x 1.871 = 0.7412 m2.At point 0.2*q f=S f . p .g(Fuel Density (p) = 800 kg / m3; g = 9.8)qf =(0.7412 x 800 x 9.8) = 5811.2

Sf fuel tank 1 at 0.8 , chord = 1.78H1 = 0.2646 m ; H2 = 0.1513 ma = Rear position spar – Front position sparFront position spar = 0.25 x 1.78 = 0.445 m.Rear position spar = 0.74 x 1.78 = 1.317 m*a = 1.317 – 0.445 = 0.872 m.*Sf fuel tank 2 at 0.8 = 0.9 x (0.2646 + 0.1513/2) x 0.8722 = 0.9 x 0.20 x 0.8722 = 0.1632 m2.

[127]

At point 0.8*qf =Sf.p.g(Fuel Density(p) = 800 kg / m3 ; g = 9.8) *qf =(0.1632 x 800 x 9.8)= 1280.7

*qfu = qf.nl.f

nl = limit load factor,is calculated according to AR-25,FAR-25,JAR-25

n ≤ 2.1 + ( 10890G+4540 )

G = Take of weight = 39110 kg

n=2.1+ 1089039110+4540

=2.3494

: taken=2.5Then f = 1.5q f u=7097.9664∗2.5∗1.5=26617.374

G fi=q fi+q fi+12

∗li

li= Length b/w two tanks, calculated below

LENGTH OF FUEL TANK CALCULATION:-

Total length of wing = 22.73 m.Semi-span = 11.363 m.

11.363−Diameter of fuselage2

=11.363−2.52=10.115m

Equivalent wing of 10.115 m is divided in 11 sections10.115

2=0.91954 m

No of section b/w 0.2 & 0.8 = 6i.e.Length=6∗0.91954=5.56172 m.

G fi=q fi+q fi+12

∗li

qfi +qfi+1 = 7097.9664

G fi=7097.966+12

∗5.561 = 19738.674

хf

аb

c

H1 H2

c.g. of fuela fuel tanka front spar

a rear spar

[128]

In a wing the fuel can be placed in space between front and rear spar in cross section and on length from root rid up to tip rib. In the transport plane from conditions of safety the fuel is forbidden for placing in a fuselage. They can be placed in a wing and tail unit. On semi span of wing they are placed from two up to five tanks. One tank is forbidden for putting, that at damage of a tank all fuel was not poured out and for observance of positioning. The heavier plane has the more tanks. If the engines are established on a wing, between the engine and tank there should be a fire-prevention interval not less 200mm. If the engines are established on engine pylon, the fuel can be placed on all semi span of wing. For simplicity of border of fuel tanks it is possible to combine with points of the task of aerodynamic load. It is possible to place fuel at once on an equivalent straight wing.

[129]

In The position of a centre of gravity for fuel in cross section хf can be estimated as the centre of gravity for a trapeze under the formula

At point 0.2

= 1.871

3×

0.3247+2(0.5680)0.3247+0.5680

= 1.02m Where хf - is distance from front spar up to a centre of gravity of fuel in the given cross section.

On this size it is possible to find relative size for a position of a centre of gravity in percentage of chord from leading edge of a wing

= 1.02+0.95

3.82×100

= 51.57 %

It is possible to consider this size as constant size on the wing span.

AIR LOAD CALCULATION

The Y air load is allocated according to the relative circulation low, i.e.

q yа (z )= n⋅f⋅g⋅M t

Lw

Г (z ) ,

z= z0,5 Lw

where Г ( z ) - is relative circulation, Mt- is the take-off weight of the plane (it is given in the initial data), n, f – limit load factor and factor of safety, Lw – wingspan.

Distribution relative circulation on wingspan straight trapezoidal center-section-less wing

Гf (5 10)2z/l = 40.0 1,3859

[130]

0.1 1,37010,2 1,32450.3 1,25240.4 1,16010.5 1,05530.6 0,94190.7 0,82710.8 0,70510.9 0,54340,95 0,40921 0

1. Wing has not center-section (2 lc = 0).

2. Wing is plane.

3. Wing aspect ratioλ = Lw

2

/Sw = 7.85

4. Wing taper is η=b0 /b t = 4

5. For low-wing monoplane Гf is given from board rib, for midwing and high-wing Гp is given from axial rib.

6. If wing taper differentiates from table data, valises Гf are calculated by linear interpolation.

The function Г ( z ) depends from many factors, from which in the given work we take into account only the dependence from wing taper and sweepback.

The Г ( z ) = Г f (z ) function values for plane straight trapezoidal center-section-less wing

are reduced in table # 2. = 45°.

2z/l ΔΓx(450) ΔΓx(x)0.0 -0.235 -0.11860.1 -0.175 -0.088310,2 -0.123 -0.06210.3 -0.072 -0.03630.4 -0.025 -0.01260.5 0.025 0.01260.6 0.073 0.0368

[131]

0.7 0.111 0.05600.8 0.135 0.06810.9 0.140 0.07070,95 0.125 0.06311 0 0

Essential influence on distribution of air loading renders a wing sweep. Relative circulation in this case is determined under the formula:

Г ( z ) = Г f (z )+ ΔΓ χ ( z )

where ΔΓs ( z ) is amendment on the wing sweep.This amendment is calculated upon formulas:

ΔΓ χ ( χ )=ΔΓ(45о ) χ

45о

where the is the designing wing sweep on the chord’s fourth, angle in degree.

THE WING STRUCTURE MASS LOAD ALLOCATION.

In approximate calculations it is possible to consider, that load per unit of length of wing mass forces is proportional to chords. Then the next formula is used:

q yw ( z )=

n f⋅Gw

Sw

⋅b (z )

where the b(z) is the wing chord.After the component calculations it is possible to compute the total distributed wing

load, acting in the direction of the axis Y in the speed coordinate system. Calculations are put into the tab.# 4. At this action the coordinates origin is put into the wing root and cross sections are enumerated from the wing root in the wing tip direction, beginning from the i

= 0. The letterZ accentuate relative coordinateZ = 2⋅z /Lw . Since on the Z = 1 − 0.95

cross sections the q diagram are moved away from a straight line, it is necessary to introduce

the cross section with the Z = 0.95 coordinate (see tab. # #2, 4). The total distributed wing load is calculated under the formula:

q=q ya−q y

w−q yf

Where q yf

– is distributed ultimate fuel load from formula (2), if you consider this load distributed and it is equal to:

[132]

q yf =q f nf .

It is also necessary to plot the q y

a

, q y

w

, q yf

and q functions in the same coordinate system and in the same scale.

i z Γ f ΔΓx Γ qay

KNm

qwKNm

b(z) q tKNm

1 2 3 4 5 6 7 8 9

0 0 1.3731 151.856 -9.073 4.5 142.783

1 0.1 1.3580 153.594 -8.387 4.16 145.207

2 0.2 1.3143 151.269 -7.702 3.82 143.567

3 0.3 1.2435 145.721 -7.016 3.48 138.705

4 0.4 1.1565 137.501 -6.331 3.14 131.17

5 0.5 1.0551 127.843 -5.645 2.80 122.198

6 0.6 0.9464 117.274 -4.839 2.40 112.435

7 0.7 0.8351 105.819 -4.274 2.12 101.545

8 0.8 0.7164 92.6502 -3.589 1.78 89.061

9 0.9 0.5630 73.585 -2.903 1.44 70.682

10 0.95 0.4242 56.594 -2.560 1.27 54.034

11 1.0 0 0 0 0 -2.217 1.1 -2.217

SHEAR FORCE, BENDING MOMENT AND REDUCED MOMENT

At calculation of the allocation law of shear forces and bending moments on the wing

length in the beginning functions Qd ( z ) and M d ( z ) affected by the distributed load q (z)

[133]

are found. For this purpose integrals are calculated by a tabulated way with trapezoids and mean-value methods.

Q = ∫Lw

2

z

q ( z )dz

,

M = ∫Lw

2

z

Q( z )dz

We shall use the following formulae for the determination of distributer load q (z), shear force Q(z) and Bending Moment Md(z).

∆Zi = (Z d̅ i - Z d̅ i−1 ¿×LW

2

∆Qi = (q d̅ i + q d̅ i−1 ¿×∆ zi

2

Qi = ∆ Q+Qi+1

∆Mi = (Q d̅ i + Q d̅ i−1¿×∆ zi

2

Mi = ∆ M+M i+1

i z m ∆ z m q iKNm

∆ Qid

KN

Qid KN ∆ M id KNM M id KNM

1 2 3 4 5 6 7 8

0 0 0 142.78 161.182 1270.162 1331.605 6055.338

1 0.1 1.1195 145.20 161.618 1108.98 1150.934 4723.739

2 0.2 1.1195 143.56 157.98 947.362 972.055 3572.798

3 0.3 1.1195 138.70 151.04 789.382 799.077 2600.75

4 0.4 1.1195 131.17 141.80 638.342 635.194 1801.673

5 0.5 1.1195 122.19 131.316 496.542 482.331 1166.479

6 0.6 1.1195 112.43 119.75 365.226 341.809 684.148

7 0.7 1.1195 101.54 106.678 245.476 215.078 342.339

8 0.8 1.1195 89.06 89.406 138.798 105.329 127.261

9 0.9 1.1195 70.68 34.893 49.392 17.876 21.932

10 0.95 1.1195 54.03 14.499 14.449 4.056 4.056

[134]

11 1.0 0 -2.21 0 0 0 0

When we consider the shear forces and bending moments resulting from concentrated mass forces, from engine, we shall note the following results in the table below.

Δ z i = (Z̄ i−Z̄ i−1)⋅Lw /2 ; z0 = 0, (i = 11, 10, 9... 1),

Q i = ΔQ i+1+Q i+1 , Q11 = 0; (i = 10, 9... 1)

Δ M iс =(Q i+Q i−1)⋅Δzi

2 , M0с =0, (i = 11, 10, 9... 1)

M iс= ΔM i+1 с+M i+1 с , M11с = 0; (i = 10, 9... 1) I z m ∆ z m ∆ Qic KN Qic KN ∆ M ic KNM M ic KNM

1 2 3 4 5 6 7

0 0 0 0 165.48 185.238 -643.5984

1 0.1 1.1195 0 165.48 185.238 -458.3604

2 0.2 1.1195 111.73 165.48 122.7030 273.1221

3 0.3 1.1195 0 53.75 60.1677 150.4191

4 0.4 1.1195 0 53.75 60.1677 90.2514

5 0.5 1.1195 53.65 53.75 30.0837 30.0837

6 0.6 1.1195 0 0 0 0

7 0.7 1.1195 0 0 0 0

8 0.8 1.1195 0 0 0 0

9 0.9 1.1195 0 0 0 0

10 0.95 1.1195 0 0 0 0

11 1.0 1.1195 0 0 0 0

The calculation scheme is given in, which includes the following values:

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Qid - is shear force from distributed loads Qic - is shear force from concentrated loads Qtot = Qid+ Qic with account signs;Mid - is bending moment from distributed loads Mic - is bending moment from concentrated loads Mtot = Mid + Mic with account signs. As a wing is calculated on strength in connected coordinate system for design cross section determination of shear forces and bending moments are carried out in this coordinate system. In connected coordinate system the t axis is directed on a chord of a wing, an axis n - is perpendicular to it.

The total Qtot (z) shear forces and the total Мtot (z) bending moment affected by all forces calculation scheme.

Summation of total shear forces and bending moments acting on the wing due to distributed and concentrated loads shall be calculated as follows.Qtvt = Qid+Qic and M tot = M id+M ic

The total shear force Qtot and total bending moment M tot affected by all forces

I z m ∆ z m ∆ Qic KN Qic KN ∆ M ic KNM M ic KNM

1 2 3 4 5 6 7

0 0 0 0 165.48 185.238 -643.5984

1 0.1 1.1195 0 165.48 185.238 -458.3604

2 0.2 1.1195 111.73 165.48 122.7030 273.1221

3 0.3 1.1195 0 53.75 60.1677 150.4191

4 0.4 1.1195 0 53.75 60.1677 90.2514

5 0.5 1.1195 53.65 53.75 30.0837 30.0837

6 0.6 1.1195 0 0 0 0

7 0.7 1.1195 0 0 0 0

8 0.8 1.1195 0 0 0 0

qw

qa

Reduced axis

e

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9 0.9 1.1195 0 0 0 0

10 0.95 1.1195 0 0 0 0

11 1.0 1.1195 0 0 0 0

DETERMINATION OF REDUCED SHEAR FORCE AND BENDING MOMENT

qna = q y

a×cos (θ−α )

cosθ

qnw = q y

w ×cos (θ−α )

cosθ

qnf = q y

f ×cos (θ−α )

cosθ

If we take angle of attack, α¿ 20

Then θ=tan−1 CxCy

Taking from aerodynamic properties of the airfoil,

θ=tan−1 0.0390.6274

= 3.557 o

Position of cross-section for reduced moments; we shall use the following formulaeM zi = qne−qnd

∆ M zid = (M ¿¿ zi+M zi−1)∆ z i /2¿M zi = ∆ M zi+1+M zi+1

Qn=Qtot

cos (θ−α )cosθ , = (1.001) Qn

Qt =Q tot

sin (θ−α )cosθ ,

M n = M tot⋅sin (θ−α )cosθ ,

M t = M tot⋅cos (θ−α )cosθ

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Reduced moment’s calculation for distributed loads

I z m ∆ z m ∆ Qic KN Qic KN ∆ M ic KNM M ic KNM

1 2 3 4 5 6 7

0 0 0 0 165.48 185.238 -643.5984

1 0.1 1.1195 0 165.48 185.238 -458.3604

2 0.2 1.1195 111.73 165.48 122.7030 273.1221

3 0.3 1.1195 0 53.75 60.1677 150.4191

4 0.4 1.1195 0 53.75 60.1677 90.2514

5 0.5 1.1195 53.65 53.75 30.0837 30.0837

6 0.6 1.1195 0 0 0 0

7 0.7 1.1195 0 0 0 0

8 0.8 1.1195 0 0 0 0

9 0.9 1.1195 0 0 0 0

10 0.95 1.1195 0 0 0 0

11 1.0 1.1195 0 0 0 0

Reduced Bending Moments as a result of concentrated forces, are calculated as follows,

M agzk = nu Gag

k

cos (θ−α )cosθ

rk

± Pagay

cos (θ−α )cos θ

rk

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CALCULATION SCHEME OF REDUCED MOMENT MOMENT FROM CONCENTRATED LOADS AND FROM ALL LOADS

i P y ay

KN

r

m

∆ M zc

KNM

M zc

KNM

M zd

KNM

M z tvt

KNM

1 2 3 4 5 6 7

0 0 0 0 308.2315 -1673.36 -1365.129

1 0 0 0 308.2315 -1486.026 -1177.794

2 111.73 1.88 210.052 308.2315 -1289.06 -980.828

3 0 0 098.1795

-1087.616 -989.436

4 0 0 098.1795

-887.078 -788.898

5 53.65 1.83 98.179598.1795

-691.125 -592.945

6 0 0 0 0 -536.269 -536.269

7 0 3 0 0 -391.588 -391.588

8 0 0 0 0 -227.238 -227.238

9 0 0 0 0 -84.417 -84.417

10 0 0 0 0 -26.734 -26.734

11 0 0 0 0 0 0

Conclusion: Thus the designed aircrafts 3-D model is done. The wings mass load and air load are found.

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DESIGN SECTION PART- 4

AIRCRAFT SYSTEMS

1. AIRCRAFT HYDRAULIC SYSTEM

2. COMPONENTS INVOLVED IN HYDRAULIC SYSTEM

3. HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFT

AIRCRAFT HYDRAULIC SYSTEMINTROUCTIONAircraft Hydraulics DefinitionHydraulic systems provide a means of remotely controlling a wide range of components by transmitting a force through a confined fluid. It is a system where liquid under pressure is used to transmit this energy. Hydraulic systems take engine power and convert it to hydraulic power by means of a hydraulic pump. This power can be distributed throughout the airplane by means of tubing that runs through the aircraft. Hydraulic power may be reconverted to mechanical power by means of an actuating cylinder, or turbine.

NEED FOR THE USE OF HYDRAULIC SYSTEM

Because hydraulics can transmit high forces rapidly and accurately along lightweight pipes of any size, shape and length, they are the prime source of power in aircraft systems such as flying controls, flaps, retractable undercarriages and wheel brakes.Modern aircrafts include many different types of subsystems. These subsystems are very closely interlinked to each other. One of these subsystems is hydraulic subsystem, which is usually used for actuating most of the mechanical subsystems, such as landing gear, flight control surfaces, weapons system etc. Thus the hydraulic system is a very essential part of

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the aircraft and its reliability and functionality are very essential to the flight worthiness of the whole aircraft.

PRINICIPLE OF HYDRAULICS

Hydraulics is based on a very simple fact of nature - you cannot compress a liquid. You can compress a gas (think about putting more and more air into a tire, the more you put in, the higher the pressure). If you're really strong you can compress a solid mass as well. But no matter how much pressure you apply onto a liquid, it isn't possible to compress it. Now if you put that liquid into a sealed system and push on it at one end, that pressure is transmitted through the liquid to the other end of the system. The pressure is not diminished.

Advantages of Hydraulic Systems

1. It is lighter in weight than alternate existing systems.2. It is dead beat, that is, there is an absence of sloppiness in its response to demands

placed on the system.3. It is reliable; either it works or doesn't.4. It can be easily maintained. 5. It is not a shock hazard; it is not much of a fire hazard. .

Devices Operated by Hydraulic Systems in Aircraft

1. Primary control boosters2. Retraction and extension of landing gear 3. Sweep back and forth of wings 4. Opening and closing doors and hatchways 5. Automatic pilot and gun turrets 6. Shock absorption systems and valve lifter systems 7. Dive, landing, speed and flap brakes 8. Pitch changing mechanism, spoilers on flaps

HYDRAULIC FLUIDThe primary purpose of hydraulic fluid is to transmit force from one place, through nonmoving hydraulic tubes, to another location. The advantage of using a liquid is its incompressibility. Except for minor friction losses as the fluid passes through the tubing, all applied force is transmitted throughout the system, according to Pascal. Thus, if a number of passages exist in a system, pressure can be distributed through all of them by means of the liquid.

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Hydraulic fluids are different and generally speaking they cannot be mixed. Manufacturers of hydraulic devices usually specify the type of liquid best suited for use with their equipment, in view of the working conditions, the service required, temperatures expected inside and outside the systems, pressures the liquid must with stand, the possibilities of corrosion, and other conditions that must be considered.A good hydraulic fluid must have the following properties:

1. Incompressible2. It flows with minimum friction3. Strong lubricating properties4. Maintain its property at high temperature5. Resistant to foaming Some of the properties and characteristics that must be considered when selecting a satisfactory liquid for a particular system are discussed based upon four primary considerations:

• Viscosity • Chemical stability • Flash point • Fire point

Viscosity

One of the most important properties of any hydraulic fluid is its viscosity. Viscosity is internal resistance to flow. A liquid such as gasoline flows easily (has a low viscosity) while a liquid such as tar flows slowly (has a high viscosity). Viscosity increases with temperature decrease. A satisfactory liquid for given hydraulic system must have enough body to give a good seal at pumps, valves, and pistons; but it must not be so thick that it offers resistance to flow, leading to power loss and higher operating temperatures. These factors will add to the load and to excessive wear of parts. A fluid that is too thin will also lead to rapid wear of moving parts, or of parts which have heavy loads. There are several types of viscometers that are used as an instrument to measure the viscosity of a hydraulic fluid used on aircraft systems. Such instruments measure the number of seconds it takes for a fixed quantity of liquid to flows through a small orifice of standard length and diameter at a specific temperature.Chemical Stability

Chemical stability is another property which is exceedingly important in selecting a hydraulic liquid. It is the liquid's ability to resist oxidation and deterioration for long periods.

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All liquids tend to undergo unfavorable chemical changes under severe operating conditions. This is the case, for example, when a system operates for a considerable period of time at high temperatures.Excessive temperatures have a great effect on the life of a liquid. It should be noted that the temperature of the liquid in the reservoir of an operating hydraulic system does not always represent a true state of operating conditions. Localized hot spots occur on bearings, gear teeth, or at the point where liquid under pressure is forced through a small orifice. Continuous passage of a liquid through these points may produce local temperatures high enough to carbonize or sludge the liquid, yet the liquid in the reservoir may not indicate an excessively high temperature. Liquids with a high viscosity have a greater resistance to heat than light or low viscosity liquids which have been derived from the same source. The average hydraulic liquid has a low viscosity. Fortunately, there is a wide choice of liquids available for use within the viscosity range required of hydraulic liquids.Flash point

Flash point is the temperature at which the fluid produces enough combustible vapor that it will ignite momentarily or flash when a flame is applied. A fluid with a high flash point can get very hot before it becomes susceptible to flashing. Looking at it from another perspective, a fluid with a high flash point has minimal evaporation under normal operating conditions; therefore, a high flash point is a desirable characteristic of hydraulic fluids.

Fire Point

This is the next step up the temperature spectrum from flash point. A hydraulic fluid’s fire point is the temperature at which a substance gives off vapor in sufficient quantity to ignite and continue to burn when exposed to a spark or flame. Like flash point, a high fire point is required for desirable hydraulic liquid’s operating condition.Precautions while using hydraulic fluid:

1. Hydraulic fluid should not be mixed2. Hydraulic fluids are flammable3. Hydraulic fluids are supposed to have adverse effects like nerve damage, hence avoid

contact with skin.Examples of Hydraulic fluid used in Aviation:

1. MIL H-5606 (red petroleum based)- first introduced over fifty years ago and still used on many aircraft. Used on business jets and many U.S. Air Force aircraft, it is highly flammable and considered responsible for the loss of at least one military aircraft, due to the fire created.

2. MIL-H-83282 - first used by the Air Force in 1982 and the U.S. Navy in 1997, it is less flammable than 5606, but much more viscous at low temperature. The lower

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temperature limit of MIL-H-83282 is considered 40° F, and it is used in virtually all Navy aircraft.

3. Skydrol (purple synthetic)- these alkyl phosphate ester based fluids are used on commercial aircraft, and are less flammable than the military fluids described above. Maximum temperature limit is 160° F. These fluids have been around at least since the 1960's.

4. MIL-H-87257 - this newest fluid is used in C135, E3, and U2 aircraft; it is less flammable than 5606 (similar to 83282) but its viscosity at low temperatures allows use down to 65° F. 

COMPONENTS INVOLVED IN HYDRAULIC SYSTEMA. hydraulic pump converts mechanical power to hydraulic power

When a hydraulic pump operates, it performs two functions. First, its mechanical action creates a vacuum at the pump inlet which allows atmospheric pressure to force liquid from the reservoir into the inlet line to the pump. Second, its mechanical action delivers this liquid to the pump outlet and forces it into the hydraulic system.

Pump Cross Section

A pump produces liquid movement or flow: it does not generate pressure. It produces the flow necessary for the development of pressure which is a function of resistance to fluid flow in the system. For example, the pressure of the fluid at the pump outlet is zero for a pump not connected to a system (load). Further, for a pump delivering into a system, the pressure will rise only to the level necessary to overcome the resistance of the load.

Types of Power Pumps

There are two types of power pumps, a gear pump and a piston pump.

A. Gear pumps have efficiencies that average about 70-80% overall efficiency, where overall efficiency is defined by (mechanical efficiency)*(volumetric efficiency)

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Gear pumps move fluid based upon the number of gear teeth and the volume spacing between gear teeth.

External-gear pumps

Gear pumps can be divided into external and internal-gear types. These pumps come with a straight spur, helical, or herringbone gears. Straight spur gears are easiest to cut and are the most widely used. Helical and herringbone gears run more quietly, but cost more.

A gear pump produces flow by carrying fluid in between the teeth of two meshing gears. One gear is driven by the drive shaft and turns the idler gear. The chambers formed between adjacent gear teeth are enclosed by the pump housing and side plates (also called wear or pressure plates).

A partial vacuum is created at the pump inlet as the gear teeth unmesh. Fluid flows in to fill the space and is carried around the outside of the gears. As the teeth mesh again at the outlet end, the fluid is forced out.

Internal-gear pumps

Internal-gear pumps, Figure 4, have an internal gear and an external gear. Because these pumps have one or two less teeth in the inner gear than the outer, relative speeds of the inner and outer gears in these designs are low. For example, if the number of teeth in the inner and outer gears were 10 and 11 respectively, the inner gear would turn 11 revolutions, while the outer would turn 10. This low relative speed means a low wear rate. These pumps are small, compact units.

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B. Piston pumps

The piston pump is a rotary unit which uses the principle of the reciprocating pump to produce fluid flow. Instead of using a single piston, these pumps have many piston-cylinder combinations. Part of the pump mechanism rotates about a drive shaft to generate the reciprocating motions, which draw fluid into each cylinder and then expels it, producing flow. There are two basic types, axial and radial piston; both area available as fixed and variable displacement pumps. The second variety often is capable of variable reversible (over center) displacement.

Most axial and radial piston pumps lend themselves to variable as well as fixed displacement designs. Variable displacement pumps tend to be somewhat larger and heavier, because they have added internal controls, such as hand wheel, electric motor, hydraulic cylinder, servo, and mechanical stem.

Axial piston pumps

The pistons in an axial piston pump reciprocate parallel to the centerline of the drive shaft of the piston block. That is, rotary shaft motion is converted into axial reciprocating motion. Most axial piston pumps are multi-piston and use check valves or port plates to direct liquid flow from inlet to discharge.

Axial-piston pump varies displacement by changing angle of swash plate.

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Inline piston pumps

The simplest type of axial piston pump is the swash plate design in which a cylinder block is turned by the drive shaft. Pistons fitted to bores in the cylinder block are connected through piston shoes and a retracting ring, so that the shoes bear against an angled swashplate.

As the block turns, Figure 8, the piston shoes follow the swash plate, causing the pistons to reciprocate. The ports are arranged in the valve plate so that the pistons pass the inlet as they are pulled out and the outlet as they are forced back in. In these pumps, displacement is determined by the size and number of pistons as well as their stroke length, which varies with the swash plate angle.

In variable displacement models of the inline pump, the swash plate swings in a movable yoke. Pivoting the yoke on a pintle changes the swash plate angle to increase or decrease the piston stroke. The yoke can be positioned with a variety of controls,i.e., manual, servo, compensator, hand wheel, etc.

Bent axis pumps

This pump consists of a drive shaft which rotates the pistons, a cylinder block, and a stationary valving surface facing the cylinder block bores which ports the inlet and outlet flow. The drive shaft axis is angular in relation to the cylinder block axis. Rotation of the drive shaft causes rotation of the pistons and the cylinder block.

Because the plane of rotation of the pistons is at an angle to the valving surface plane, the distance between any one of the pistons and the valving surface continually changes during rotation. Each individual piston moves away from the valving surface during one-half of the shaft revolution and toward the valving surface during the other half.

Radial-piston pumps

In these pumps, the pistons are arranged radially in a cylinder block; they move perpendicularly to the shaft centerline. Two basic types are available: one uses cylindrically shaped pistons, the other ball pistons. They may also be classified according to the porting arrangement: check valve or pintle valve. They are available in fixed and variable displacement, and variable reversible (over-center) displacement.

In pintle-ported radial piston pump, Figure 9, the cylinder block rotates on a stationary pintle and inside a circular reacting ring or rotor. As the block rotates, centrifugal force, charging pressure, or some form of mechanical action causes the pistons to follow the inner surface of the ring, which is offset from the centerline of the cylinder block. As the pistons reciprocate

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in their bores, porting in the pintle permits them to take in fluid as they move outward and discharge it as they move in.

Plunger pumps

These reciprocating pumps are somewhat similar to rotary piston types, in that pumping is the result of pistons reciprocating in cylinder bores. However, the cylinders are fixed in these pumps; they do not rotate around the drive shaft. Pistons may be reciprocated by a crankshaft, by eccentrics on a shaft, or by a wobble plate. When eccentrics are used, return stroke is by springs. Because valving cannot be supplied by covering and uncovering ports as rotation occurs, inlet and outlet check valves may be used in these pumps.

Because of their construction, these pumps offer two features other pumps do not have: one has a more positive sealing between inlet and outlet, permitting higher pressures without excessive leakage of slip. The other is that in many pumps, lubrication of moving parts other than the piston and cylindrical bore may be independent of the liquid being pumped. Therefore, liquids with poor lubricating properties can be pumped. Volumetric and overall efficiencies are close to those of axial and radial piston pumps.

B. An actuating cylinder (Hydraulic Actuators) converts hydraulic power to mechanical power

Hydraulic actuators are devices for converting hydraulic pressure to mechanical motion (work). The most commonly utilized actuator is actuating cylinder; however, servo actuators and hydraulic motors are also employed for special applications where modified motion is required. Actuating cylinders are used for direct and positive movement such as retracting and extending of landing gear, wing flaps, spoilers and slats.

Balanced Actuator Schematic

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Design of actuating cylinders is determined by the functions that they are to perform. Actuating cylinders can be a single-acting or a double-acting cylinder types. In case of a single-acting cylinder, hydraulic pressure is applied to one side of the piston to provide force in one direction only. When hydraulic pressure is removed from the piston, a return spring moves the piston to its start position. Where as a double acting cylinder is designed so that hydraulic pressure can be applied to both sides of the piston. Thus, the cylinder can provide force in either direction. So, double acting cylinders are widely used for the operation of retractable landing gear, wing flaps, spoilers, bus doors and other similar applications.

Unbalanced Actuator

Servo actuators are designed to provide hydraulic power to aid the pilot in the movement of various aircraft controls. Such actuators usually include an actuating cylinder, a multi-port flow control valve, check valves, and relief valves together with connecting linkages. Servo actuators are employed in situations where accurately controlled intermediate positions of units are required. The servo unit feeds back position information to the pilot’s control, thus making it possible for the pilot to select any control position required. Such actuators are used to move large control surfaces such as aircraft rudder, elevator, and ailerons. For example, servo units are used to aid the pilot in the operation of collective pitch and throttle control lever in Mi-24/35 helicopter’s friction clutch.

C. Hydraulic System Accumulators

Purpose of Accumulators in the System: 1. Absorbs the shock due to rapid pressure variations in a hydraulic system 2. Helps to maintain a constant pressure within the hydraulic system 3. Helps the hydraulic pump under peak pressure loads 4. It is an emergency source of power (the braking system has its own accumulator)

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Piston Accumulator Schematic Metal Bellows Accumulator Diaphragm AccumulatorPrinciple of Operation

At the bottom of the accumulator is a gas valve. Compressed gas at about one half the system pressure is let into the accumulator through the gas valve. This forces the diaphragm that separates the oil side from the gas side to "pop" up towards the oil side.Then oil is sent through the system. When the system pressure reaches a point when it is greater than the pressure of the accumulator, the diaphragm will deploy (inflate). Using Boyle’s Law, the compressed gas will increase in pressure as its volume decreases. The diaphragm will move up or down, depending on system pressure. When the diaphragm is at half way, the gas volume will be ½ as much as it was initially, while the accumulator pressure will be twice as much as its pre-load pressure (i.e., 1/2 system pressure). Therefore when the accumulator is at half volume of gas, it will be charged at full system pressure.Accumulator Shapes Accumulator can have different shapes according to its purpose. The shape can be; spherical or cylindrical or bottle type. The spherical shape is the strongest and effective single shell body used to withstand high pressure before failing. The bottle type accumulator is not widely used on most aircrafts because of the effect on bladder used to expand and contract. Also the cylindrical type is not used very often because the friction will cause wearing of the body and piston, thereby allowing the gas pressure to escape.

D.Distribution Devices and Reducers

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Distribution devices (valves) are the main part of hydraulic system to control a flow of fluid to the actuating mechanisms in aircraft control system. These hydraulic valves may be classified according to the following features:

Drive type: manual or electrically (solenoid) controlled Number of fixed positions of the wing: two or three-position valves Type of distribution device: slide valve or plug type Control method: direct control or servo control method

Hydraulic reducers are intended to decrease the fluid pressure on separate portions of the system to the required value. They are necessitated owing to the fact that aircraft hydraulic systems use a number of actuating devices which are operable according to the specifications on lower pressure than the working pressure in the supply line. The hydraulic reducers can be of a constant-pressure reducer which maintains a definite fluid pressure or variable-pressure reducers in which the pressure is set in the course of control.

E. Pressure Boosters and De-boosters Pressure Boosters Pressure boosters are rarely used in aircraft (almost all planes use de-boosters). If we need higher pressure, we must change the entire power and actuating system. This adds weight that is not needed. In general, we cannot put-in larger pistons and piston cylinders to increase the power, because, normally we don’t have the room for it. A simple solution is to raise the pressure in a localized area.The function of a pressure booster is to act like a transformer; i.e, it raises the pressure of a small circuit connected to the power system. The booster is a cylinder made up of two pistons of different surface areas that are connected. The larger surface area (A1) is connected to the inlet side of the hydraulic system, and, the smaller surface area (A2) is connected to the outlet side of the hydraulic system as shown in the figure below.

Pressure booster’s operationDisadvantages of a Pressure Booster

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Weight of the pressure booster is high if it is put into the aircraft, thereby reducing payload that the aircraft can carry.

It requires a very large booster stroke to meet the requirement Pressure boosters must be built into the aircraft during the aircraft's construction, if

high pressure is needed in the hydraulic system. Leakage from the pressure booster is an important factor and will increase

possibility of fire hazards. Pressure De-Boosters Pressure de-boosters are used to reduce the pressure in the system to a level that can be used by certain devices. Pressure de-boosters are pressure boosters turned upside down (that is, the inlet side of the booster has the smaller area piston and the outlet side has the larger area piston). They are employed in power brake systems, using engine power to help apply the brakes. Since aircraft wheels are made of magnesium, any high pressure on the wheels will cause them to split. That is why they must de-boost the pressure gotten from engine.

The inlet line to the de-booster comes from the power brake control valve. The outlet line goes to the brake system. This valve meters hydraulic fluid and hydraulic pressure directly. The force applied to the wheels to make them stop is proportional to how hard you push on the rudder pedal. It is normally used for a 1 or 2 cubic inch application.

Pressure de-booster’s operation

Basic Operations in Hydraulic Controls

Pressure Control (Pressure limiting devices/relief valves) Function Pressure limiting device or relief valves are required to limit the pressure in some section of the hydraulic system to a predetermined level. That pressure level may be considered dangerous and, therefore, must be limited in such a case.

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Principle of Operation

The adjustment screw at the top of the pressure relief valve is set for a certain pressure value, let us call it P2. In general, even with a pressure of P1, the poppet would lift up, except that the spring is strong and has downward force forcing the poppet closed. Poppet will not move until a pressure greater than that required is felt by the system (i.e., P1>P2).

Pressure limiting device operating principle

When the pressure increases, the poppet will move up, forcing the excess liquid to move through opening at high velocity. On other side of seat, pressure is zero because the back side of the relief valve is connected to the return line. When the pressure in the system decreases below maximum, poppet will return to its seated position, sealing the orifice and allowing the fluid to follow its normal path. These type of pressure relief valves are only made to be used intermittently.Flow Control Selector Valves Selector valves are used as (1) directional control devices to insure the movement of hydraulic fluid flow in the proper direction, and (2) as stop-locks to lock the selector switch in a certain position. Types There are three types of selector valves used in aircraft hydraulic system namely; rotary type, piston type and poppet type. Rotary types Rotary type selector valves are plugs within which are passage ways for the fluid to move through. Tubing from the hydraulic pump or return line are connected to the rest of the hydraulic system by movement of the plug. You cannot use high pressure oil because of leakage around the plug. To reduce leakage, you might want to make the plug fit more tightly into the selector valve body. However, the better you make the fit, the more friction will exist between the plug and the selector valve body, making it difficult to operate.

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Rotary type selector valvePiston Type

Positions 1, 2 and 3 (shown in fig.--) are representative positions for the piston-type selector valve. Position (1) is the position of the selector valve, for example, upon the extension of landing gear or the lowering of flaps. Position (2) is the position of the selector valve upon retraction of the landing gear or the raising of the flaps. Position (3) is the stop-locking position of this type of valve. This piston type valve uses the Vickers spool mechanism in which the piston "lands" isolate the high pressure oil (red area) from the low pressure oil (blue area).

Piston type selector valves operating principlePoppet Type – Stacked Poppet In this type of valve, any movement of the handle (at the lower right of the diagram) changes the camshaft and cam settings, thereby opening and closing the poppet valves and letting

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high and low pressure oil to the proper sides of the actuating cylinder and return line, respectively.These mechanical type selector valves require a fair amount of tubing. In order to reduce the amount of tubing, electric switches have been used to operate solenoids which operate the selector valves. This has the added advantage of reducing the wasteful motions of pilot. This type of combined electronic circuits and hydraulic system is called electro-hydraulics.

Stacked poppet type selector valves operating principle

Flow Restrictors Since the speed of the actuating cylinder is determined by the rate of flow of the hydraulic fluid, we may need a device to control the rate of flow. This device is called a flow restrictor. Since none of the selector valves meter the flow, we must use the restrictor. There are four types of restrictors used in aircrafts hydraulic system:

1. One way fixed restrictor 2. One way variable restrictor 3. Two way fixed restrictor 4. Two way variable restrictor

One Way Fixed Restrictor The One Way Fixed Restrictor is not used all the time, but, it is being used more than the other types of restrictors. It is a check valve type restrictor with a drilled hole through the seat to the other side of the check valve. When the flow pressure seats the check valve ball (i.e., flow moving from right to left), some of the fluid can still reach the other side through the drilled hole in the seat. However, since the hole size is fixed, the amount of fluid passing through the passage to the other side is also fixed.

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One way fixed restrictorTwo Way Fixed Restrictor The Two Way Fixed Restrictor is not used because it restricts the flow on the side of the restrictor where we want the flow to occur normally. Because the passage size is fixed, the amount of fluid moving from right to left, or vice versa, is fixed, as well.

Two way fixed restrictorOne Way Adjustable Restrictor The One Way Adjustable Restrictor is being used nowadays. It is the same as the One Way Fixed Restrictor but the amount of fluid passed through the drilled opening in the seat is regulated by means of an adjustment screw.

One way adjustable restrictorTwo-way Adjustable Restrictor The Two Way Adjustable Restrictor is the same as the Two Way Fixed Restrictor, but it also has an adjustable screw that can be used to further restrict the amount of hydraulic fluid passing through the opening.

Two way adjustable restrictorIf a system relief valve (SRV) were used to regulate pressure, it would have to be replaced in a very short time. This would be due to the overuse of the SRV and the failure of the

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spring's elasticity. If the SRV were used, the oil pushing on the spring-ball combination would cause tremendous vibrations and heat would be dissipated by the oil under high pressure attempting to push the ball away from the seat to get to the low pressure side. The range of operation of pressure regulator is defined by the difference in force required for bypass and the force required at actuation. And the dual purpose of this pressure regulator is to reduce the load on the hydraulic pump when not needed and to keep the hydraulic pressure within the operating range of the hydraulic system.

Douglass Pressure Regulator

When an actuating cylinder finishes its motion and stops, a high pressure will be felt through the system. If so, this high pressure oil coming from the power pump (right side of diagram) will keep check valve C open and also act on piston A. In its movement, piston A pushes Ball B off seat D. The oil, taking the passage of least resistance, goes through passage D into the center chamber (colored blue) back to the reservoir. The pressure on the right side of check valve C will drop and will be less than the pressure on the left side of C, therefore, causing the ball to seat itself in check valve C. When the hydraulic system pressure drops, the pressure on piston A decreases, causing a decrease in pressure on B as well. The path of least resistance through D will close and the oil will move in the direction towards check valve C. Now, because the pressure on the right side of C is greater than on the left of C, the check valve will be forced to open and the oil will move toward the selector valve side of the system (left side of diagram).

HYDRAULIC DESCRIPTION OF THE DESIGNED AIRCRAFTThe airplane is equipped with two independent hydraulic systems, each powered by one engine driven-pump and one electric motor-driven pump.

Hydraulic power is provided by three independent systems designated No.1, No.2 and No.3.All systems operate at a nominal pressure of 3000 psi (20,685 kPa)

Hydraulic fluid “Skydrol”- Skydrol 500B-4 (Type IV class 2): The Skydrol 500 series of fluids has the longest service history among phosphate ester products. The first version, Skydrol 500, was introduced in 1952. Steady improvements to the formulation led in 1978 to the current version, Skydrol 500B-4 which contains the same breakthrough anti erosion additive and acid scavenger found in Skydrol LD-4 . Skydrol 500B-4 is the most worker

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friendly of the aviation phosphate esters; it is least irritating to skin and less prone to form mists which can be irritating to the respiratory tract. This has given the product enormous popularity for use in workshops and indoor test stands.

Specification of Skydrol

Property Skydrol® 500B-4Appearance Clear, purple liquidNeutralization number 0.10 max.Moisture content (wt.%) 20 maxSpecific gravity, 77 °F/77 °F 1.050-1.062Viscosity in cSt at 210 °F/99 °C 3.68-4.00at 100 °F/38 °C 11.4-12.4at -65 °F/-54 °C 4200 max.Flash point, COC, 350 °F/177 °CRefractive Index @ 25 °C 466 to 1.474

Skydrol is highly corrosive and can produce severe skin and eye irritation.

SYSTEM DESCRIPTION

Each hydraulic system consists of a hydraulic fluid reservoir, a manifold, one engine-driven pump, one electric motor-driven pump, one shutoff valve, one accumulator and a priority valve installed in the hydraulic system 1.

RESERVOIRThe hydraulic fluid stored in the reservoir is pressurized, to avoid pump cavitations. This pressurization function is performed by fluid drained from the pressure line. The reservoir is equipped with a quantityIndicator which transmits information to the MFD and EICAS displays for indication and warning purposes. A thermal switch is responsible for the high temperature message, if the fluid temperature increases above 90°C.

SHUTOFF VALVEA shutoff valve is installed between the reservoir and the engine-driven pump. It cuts the hydraulic fluid supply to the engine-driven pump, if there is a fire on the related engine or in case of hydraulic fluid overheats. This valve may be closed either through the engine fire extinguishing handle or through a dedicated button on the overhead panel.

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ENGINE-DRIVEN PUMPThe engine-driven pump provides continuous fluid flow at 3000 psi for operation of the various airplane hydraulically-powered systems. The pump is connected to the engine accessory gearbox and, as long as engine is running, it generates hydraulic pressure. During engine start the fluid remaining in the suction line is sufficient to avoid pump cavitation and provide reservoir pressurization.

ELECTRIC MOTOR-DRIVEN PUMPThe electric motor-driven pump has the same connections as the engine-driven pump, but has a lower flow capacity. The pump normally operates in the automatic setting mode, turning on when the associated hydraulic pressure drops below 1600 psi or the associated engine N2 drops below 56.4%. If the pump starts operating in the automatic mode, it will be turned off after the pressure or N2 are reestablished to normal values. The electric pump may be turned on at pilot command, through the selector knob on the overhead panel, furnishing continuous fluid flow at 2900 psi.

MANIFOLDThe manifold provides the following functions:-Fluid filtering (pressure and return lines).-Overpressure relief (main and electrical pumps).-Pressure indications (main and electrical pumps).Fluid leaving the pump flows to the manifold, where it is filtered and then routed to the airplane systems. Inside the manifold, a check valve prevents the fluid from returning to the pump, while a relief valvediverts the excess fluid to the return line. The return line is supplied by the fluid coming from the airplane systems, fluid drained from the pump, fluid from the relief valve, and fluid refilled by the maintenance personnel. Under any situation the fluid is filtered and returned to the reservoir. The manifold incorporates two pressure switches to detect low hydraulic pressure, and a pressure transducer to indicate system pressure. Signals from the pressure switches and pressure transducer are sent to the MFD and EICAS displays.

PRIORITY VALVEThe hydraulic system 1 incorporates a priority valve. If the system is powered by the electric motor-driven pump and the landing gear is commanded to retract, the valve will provide minimum flow to theLanding gear system and give priority to the flight control services. In this case, the landing gear will operate through the accumulator pressure.

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ACCUMULATOREach hydraulic system has one accumulator. The function of the accumulator is to keep the surges of the hydraulic pumps at a minimum, and to keep a 3000 psi pressure available for operation of the landing gear and main door (system 1) or operation of the emergency parking brake (system 2).

EICAS MESSAGESTYPE MESSAGE MEANINGCAUTION HYD SYS 1 (2) FAIL

HYD SYS 1 (2) OVHT

Associated hydraulic systemis not pressurized (inhibitedwhen the airplane is on theground, engine is shut down and parking brake is applied

Associated hydraulic systemfluid temperature is above90C.

ADVISORY E1 (2) HYD PUMP FAIL

E1 (2) HYDSOV CLSD

HYD1 (2) LO QTY

HYD PUMP SELEC OFF

Engine-driven pump is notgenerating pressure with associated engine running.

Associated hydraulic shutoffvalve is closed.

Fluid level in the associatedreservoir is below one liter.Report to the maintenancepersonnel if the hydraulicreservoir operates empty.

Associated electric pumpselected OFF with theparking brake released.

CONTROLS AND INDICATORS

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HYDRAULIC SYSTEM PANEL

1- ENGINE PUMP SHUTOFF BUTTON (guarded) Closes (pressed) or opens (released) the associated engine pump shutoff valve. A striped bar illuminates in the button to indicate that it is pressed.

2- ELECTRIC HYDRAULIC PUMP CONTROL KNOB

OFF - Associated pump is turned off.

AUTO - Associated pump is kept in standby mode, ready to operate if the engine-driven pump outlet pressure drops below 1600 psi or the associated engine N2 drops below 56.4%.

ON - Associated pump is turned on.HYDRAULIC PAGE ON MFD

1- FLUID QUANTITY INDICATION

Ranges from zero to maximum hydraulic fluid quantity. Scale (horizontal line) and pointer: green when greater than 1 liter. amber when equal to or less than 1 liter. Pointer disappears if data is invalid.

2- PRESSURE INDICATION

Ranges from 0 to 4000 psi, with a resolution of 100 psi. Digits: green from 1300 to 3300 psi. amber and boxed below 1300 and above 3300 psi. Digits are replaced by amber dashes if data is invalid.

3- ELECTRIC PUMP STATUS Indicated by the green label ON or OFF.

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HYDRAULIC SYSTEM MAINTENANCEMAINTAINENCE

The set of action including inspection, servicing, and determination of condition required to achieve a derived outcome which restore an A/C part and equipment in serviceable condition.

AIRWORTHINESSThe continuing capability of the A/C to perform in satisfactory manner, the flight

operation for which it is designed.

INSPECTIONIt is the most important form of function of aviation maintenance. As the A/C gives

complexity, it becomes more important to detect any possible trouble before it becomes serious. To assist this, aero engineers are provided with detail special check list and the maintenance manual for each type of A/C. The engineer has to go through maintenance manual thoroughly before attempting any kind of activity in aircraft and its components. The operations may be carried out on A/C on daily flying hours and/or cycle basis.

OVERHAULOverhaul means stripping a unit and restoring it to its design performance level after

replacing, reworking of parts to a given standard.

SERVICINGIt means preparing the A/C for flight, includes providing the A/C with fuel and other

fluid and gases but do not include any work that is maintenance.

TROUBLE SHOOTIt means to analyses and identify the malfunction.

REPAIRIt means to correct the defective condition.

MODIFICATIONIt is a continuous process to improve its reliability and performance.

SERVICING SCHEDULESServicing on Hours/Calendar/Cycle basis, which are to be carried out on aircraft at set Hours/Calendar/Cycle basis are mentioned in the manual or A/C servicing schedule. The

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servicing includes examination, inspection, lubrication and removal of major components such as landing gear jacks, door locks, air-conditioning equipments; aircraft brake units wheels etc. landing gear functional test, flying control range and moment check., A/C rigging procedure, hydraulic fluid contamination test, fuel contamination test & some activities requires replacement of components.

Aircraft maintenance checks are periodical checks that have to be done on all aircraft after a certain amount of time usage. Aircrafts usually refer to as one of the following checks.

A CHECKThis is performed approximately every month. This is usually done over night. The

actual occurrence of this check varies by the type, cycle or number of hours flown since the last check. The occurrence can be delayed by the aircraft if certain predetermine conditions are met.

B CHECKThis is performed in approximately 3 months.

C CHECKThis is performed every 12 to 18 months. This check puts aircraft out of service and

requires plenty of space usually at the hanger and maintenance base. Schedule and occurrence has many factors. The component is described and thus varies with the A/C category and type.

D CHECKThis is the heaviest check of an A/C. This check is done approximately every 4 to 5

year. This is the check that takes the entire A/C apart for inspection. A comprehensive check, analysis Non Destructive Testing (NDT) check and complete health monitoring of the engine has to be recorded. Complete overhauling of the A/C and its components even A/C painting is also required in this process.

AIRCRAFT GENERAL MAINTENANCE

Before caring out any work on the A/C, the respective maintenance manual is to be referred for further instructions. The necessary safety precautions are to be strictly followed. Before switching on the master battery switch ensure that the under carriage selector

lever is in down position and latched and all the armament store door switches are in safe condition.

Ensure that the wheel chocks are engaged.

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Before operating the control surface, ensure that the control locks are removed. Before starting the engine. Chocks are to be kept in front of the wheel. A/C brake system in serviceable. A serviceable fire extinguisher is available. Never tow an aircraft without a person inside the cockpit before towing the A/C,

check the brake pressure. While towing the A/C never exceed the walking speed. Never drop any tool while working. While working inside the A/C, collect all the tools and space on completion of the job

and ensure no items are left behind.

DESIGNED AIRCRAFT HYDRAULIC SYSTEM MAINTENANCE

Always release the system pressure before removing a component from the A/C Never does any maintenance work on airplane with any other specified oil other than

the recommended one. Carry out the patch test on the system to prevent the contamination of oil. This can be

carried out using Millipore patch test kit. Never mix different grade of hydraulic oil to service the A/C. Blank all the ports of the removed components and the A/C pipe ends to avoid the

entry of dust, dirt and foreign particles. Follow the necessary precautions to dismantle the hydraulic components. Avoid spilling of hydraulic fluid on the A/C and in and around from the A/C. If

spilled it should be cleaned immediately to avoid slipping. Before fitting a new hydraulic component, it should be unblanked, degreased, washed

and flushed. While fitting the non return valve and restrictors, ensure that the marked arrows are in

the desired direction.

Conclusion: The general hydraulic components are briefed and the designed aircraft hydraulic system descriptions are explained with the schematic diagram.

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TECHNOLOGICAL SECTION

INTRODUCTION TO RIB MANUFACTURE AND ASSEMBLY

The main objective of this report is to study and understand the manufacturing process and assembly scheme of the designed unit. Also the assembly base and installation base of each part is briefed which is very useful during the assembly of the rib. Some basics concepts of jig and assembly jigs are explained in this report. Basics of aircraft rib its type, rib loads, parts of rib, rib construction is discussed

The assembly process is the process by which an item or product is put together. The assembly process may include soldering, wiring, press fitting, brazing, shrink fitting, welding, adhesive bonding, riveting, and mechanical fastening. Within each of these assembly processes a series of sequences is required to accomplish the process, without regard to the product configuration, material, or quantity to be produced, or the rate of production. For example, many of the steps in creating a circuit card assembly, a wire harness, the frame of a truck, or in the installation of fittings on sailboat are essentially the same. The detailed instructions for the sequence should spell out the differences peculiar to the product at hand.

An extremely important part of the assembly process documentation is visual aids. Visual aids can be anything from an actual mockup of the product to a black and whiter color photograph, to a three-dimensional isometric or exploded-view drawing, to simple sketch, or to a tracing lifted directly from the engineering drawing

In this report we are going to see about the aircraft vertical fin ribs manufacturing process, its assembly process, and jigs used to assemble the rib, assembly diagram, exploded view diagram, basics of rib construction and rib parts.

AIRCRAFT RIB

In the framework of a wing, ribs are the crosspieces running from the leading edge to the trailing edge of the wing. The ribs give the wing its contour and shape and transmit the load from the skin to the spars. Ribs are also used in ailerons, elevators, fins, and stabilizers.

Ribs typically form part of a structure that supports and defines the shape of an aero foil surface. The rib comprises a generally planar web structure disposed between a series of rib feet associated with the upper wing skin and a series of rib feet associated with the lower wing skin.

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TYPES OF RIB:Two main classification of ribs are explained below,

COMPRESSION RIBS:Carry main load in the direction of the flight, it has firm from leading edge to the trailing edge. In some aircrafts the compression ribs is a structural piece of tubing separating two main spars. The main function of this rib is to absorb the force applied to the spar when aircraft is in flight.FORMER RIBS: It is made from light metal which is attached to the stringers and wing skins to give The wing aerodynamic shape. Nose ribs Trailing edge ribs Mid ribs running fore and aft, between front and rear spar.

The rib is mainly subjected to three kinds of loading.a) Aerodynamic loads transmitted from the skin-stringer wing panels. b) Concentrated forces transmitted to the rib due to landing gear connections, power plant’s nacelle connections, etc.

RIB PARTS, ITS USES AND LOADS ACTING ON RIB PARTSThe Rib consists of following parts RIB CAPS (RIGHT AND LEFT CAP)USES

– Rib caps are nothing but the flange of a beam– Caps enable the rib to hold tightly from top and bottom and it is also a load

carrying element. There are different types of cap profiles used in rib.Some of the types of cap are given below.

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LOADS ACTING ON CAPS• Rib caps carry the bending load as axial load.• Rib caps are designed to have maximum radius of gyration.• High local crippling stress.

RIB CAP MATERIAL

Aluminum alloy 7075 (Aluminum & zinc) - High mechanical properties and improved stress corrosion cracking resistance.

RIB WEBUSES

• Web of the rib is used to define and produce the air foil shape. It is very long, thin and flexible. Since it gives main shape to the rib, the contour of

web should be very exact and to be produced very carefully.

LOADS• To Carry inertial loads (fuel, equipment, missiles, rockets).• To Support skin-stringer panel in compression and tension.• Crushing loads due to wing bending.

WEB MATERIAL

Aluminum alloy 7075 (Aluminum & zinc) - High mechanical properties and improved stress corrosion cracking resistance.

WEB STIFFNERStiffeners are mainly substituted for the corrugated aluminum sheets. They are responsible to transfer the loads or even provide supports for the web and to the caps.

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STIFFENER LOADS Prevent the overall instability of the web. Increase the buckling strength of web.

MATERIALS Stiffener - Same material (Aluminum alloy 7075) used for web of ribs. Ductile

materials are best suited for the pressing like aluminum, mild steelLIGHTENING HOLESLightening holes are found in web of ribs. They’re purposely made in different shapes. But most common shape is circular and elliptical in some cases.

USES It is provided to stiffen the rib web and reinforce the hole edges.

LIGHTENING HOLES LOADS• Lightening holes may be introduced to the web of the rib for mass reduction.• It is also an accessibility and to form a passage for wiring and fuel pipes.

WEB FASTNERSThe fasteners we use in our rib construction includes rivets and boltsSome of the aluminums alloy rivets used in rib construction is 2017, 2024 and 2117.ATTACHEMENT PLATEUSES Provides special support to the whole structure. Increases stiffness of the web at leading edge. Also provides rigidity to the web.

MATERIAL

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Attachment plate - Aluminum alloy 7075

RIVET HOLESUSES Rivet holes are provided for installation of the rivets. Rivets are permanent mechanical fasteners to join the web along with the caps and

stiffener.MATERIALS

Rivets - Rivets of pure aluminum are used for riveting nonstructural parts fabricated using the softer aluminum alloys.

RIB CONSTRUCTION

Former ribs, located at frequent intervals throughout the wing, are made of formed sheet metal and are very lightweight. The bent-up portion of a former rib is the flange and the vertical portion is the web. The latter is generally made with beads pressed between the lightening holes. These holes lessen the rib's weight without decreasing its strength. Lightening hole area rigidity is ensured by flanging the edges of the holes. Basic rib construction is given in a figure below

The reinforced rib is similar in construction to the spar, consisting of upper and lower cap strips joined by a web plate. Vertical and diagonal angles between the cap strips reinforce the web plate. The reinforced rib is used more frequently than the truss rib.

Vertical and diagonal cross members only are used to reinforce and join the cap strips in constructing truss ribs. These and reinforced ribs are heavier than former ribs and are used only at points where the greatest stresses are imposed.

ASSEMBLY METHOD BY ASSEMBLING HOLESIt is used for flat form and simple curvature units and panels. In this method mutual position of the assembled parts is determined by coincidence of the coordinated Assembly hole.

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Which in advance are done with the help of coordinated template. The basing on the Aircraft is possible at assembly of Aircraft and Helicopter framework and skin part as well as of the product inner equipment objects, when a demanded accuracy of the assembly not exceeds by -1.5 to +1.5mm

LOFT TEMPLATE METHODDistinctive feature of the A and H manufacture is the LTM of AT manufacturing of many parts and objects of assembly by a way of dependent on sizes and forms by carrying them on standard rigging, then on a working rigging and further on products by various ways of copying. Three main scheme of coordination process includes,

1) The scheme using flat special rigid carriers as basic means- templates received by theoretical and constructive lofts.

2) The scheme of special volumetric carriers of forms and size- of standards, of mock-up made by construction method.

3) The scheme based on standard assembly unit and technological machine as a whole, received as a result of monitoring assemblies and coordination development of nodal complete set of parts.

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At LTM realization as the form and sizes standard take theoretical loft, representing the full scale drawing of the A and H aggregate, made in three projections on the rigid basis i.e.,

The lateral projection Loft The plane projection The combined cross-section Loft.

On TL the basic templates are made, which bear all necessary information for industrial templates manufacturing, and on their basis devices for parts manufacturing and AU assembling are made. CO-ORDINATE TEMPLATE METHODIn Coordinate template method the task of accuracy is increased and labor input is reduced at assembly devices coordination and coordination of their clampers was solved in LTM system by its addition of special flat and spatial coordinate benches. The Loft Conductors and Instrument Stands which are universal means for exact construction of the sizes coordinate at mounting assembly devices. So coordinate template method of coordination was originated.

Figure 1MANUFACTURING METHOD

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The rib is machined from a solid cuboidal one-piece block of metal material, known as a billet, in order to provide strength and to remove the problems associated with joining components that are made separately.

Some of the manufacturing process is given below,

The forming processes modify metal or work piece by deforming the object, that is, without removing any material. Forming is done with heat and pressure, or with mechanical force, or both.

Stamping includes a variety of sheet-metal forming manufacturing processes, such as punching using a machine press or stamping press, blanking, embossing, bending, flanging, and coining. This could be a single stage operation where every stroke of the press produce the desired form on the sheet metal part, or could occur through a series of stages. The process is usually carried out on sheet metal, but can also be used on other materials, such as polystyrene.A punch press is a type of machine press used to cut holes in material. It can be small and manually operated and hold one simple die set, or be very large, CNC operated, with a multi-station turret and hold a much larger and complex die set.Blanking and piercing both are shearing process in which a punch and die are used to modify webs. The tooling and processes are the same for two, only the terminology is different. In blanking the punched out piece is called blank. In piercing punched out piece is called scrap.Bending is a manufacturing process that produces a V-shape, U-shape, or channel shape along a straight axis in ductile materials, most commonly metal. Commonly used equipment includes box and pan brakes, brake presses, and other specialized machine presses. Typical products that are made like this are boxes such as electrical enclosures and rectangular ductwork.

PRODUCTION METHOD OF PARTS OF THE RIBTemplate of the web of rib is made first with perfect contour and then original sheet is made manually.

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RIB WEBWeb of the rib is produced by forming with hydro press, punch press with special milling device. Since the contour of our web is very complex first the template is made with marking each dimension clearly and then the manufacturing is processed if we are making it manually.

The rib web is formed from metal that lies near the surface on a first side of the billet, and the ends of the rib feet are formed from metal near the surface on the opposite side. Thus a relatively small amount of machining is conducted on the first side of the billet to form the web. Such a rib may suffer from distortion problems, which have to be accounted for during the rib machining process because the metal at and close to the surfaces of the billet is subject to residual stresses (resulting from the process used to manufacture the billet) that may cause deformation of, or undesirable internal stresses in, the web produced there from be damaged.

Finally completing the web plate make riveting holes using drill bits for exact dimensions.

STIFFENERStiffeners are made by bending process. We know that our stiffener is a L-type

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Suitable length of the aluminum stiffener is taken .Dimensions are marked and the rectangular shape is made by cutting unnecessary parts and then it bend using the bending tool.We can also use press brake forming, a work piece is positioned over the die block and the die and pressed.The L-punch forms a L-shape with a single punch.Finally completing the five stiffener make riveting holes using drill bits for exact dimensions.

MAKING THE LIGHTENING HOLESLightening holes are done by drill-bits and flange is made by form blocks. The dimensions of four holes are marked out clearly and then it is processed.The first and last hole has different dimensions, the second and third are similar in dimensions.Drill-bit is a cutting tool used to make cylindrical holes. Bits are held in a tool called grill, which rotates them and provide torque and axil force to create holes.

Flanging the Lightening Holes

Most of the rib lightening holes has a 3/8" wide flange at 30 degrees around each hole. This is to stiffen the rib web and reinforce the hole edges. There are several ways to accomplish this. Another method is using form blocks. Two large holes were cut in each of the two form blocks that were 3/4" larger than the lightening holes.

Rib was clamped between the two form blocks and bolted together through the jig holes. It was also clamped in the center between the holes to assure a nice tight squeeze. The pieces are then placed into the bench top press and the plug is then pressed down against the rib (bevel side down).

RIGHT, LEFTAND SIDE CAPS

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The three caps are manufactured using milling process, with the help of milling cutters. The right and left caps runs along the length of the rib and it has the same contour of the rib. So while making it the dimensions are marked out clearly first. The side cap which is a T-profile type attaches to the trailing edge of the rib and special holes are made there using exact dimensions.Finally after completing the caps make riveting holes.RIVETTING HOLESRivet holes for each of the part like web, top and bottom caps, and stiffener are made using a drill bits. Before riveting make sure to mark the dimensions for all rivets.

ATTACHMENT PLATEMark out the dimensions clearly first, then cut out the remaining parts using cutting tool and using milling tool the curved surface is made and required holes are placed using drill bit.

ASSEMBLY PROCEDURE OF THE RIBSCHEME OF ASSEMBLY OF THE FIN RIB

Depending on a degree of partition of the airframe and airborne systems in the assembly unit and a degree of differentiation of the assembly mounting and attachment works on the object, the assembly mounting works can be classified into three typical schemes of assembly they are

1.The sequential scheme 2. The parallel scheme 3. The parallel sequential schemeThe Sequential Scheme: This scheme is used for construction of the aircraft's aggregates in condition of the small series or individual production, when the partition scheme doesn't distinguish the panels. The parts and small assembly units are based sequentially on the basic part or on the basic unit. After that the sections, compartments, aggregates are assembled in a sequence which are joined into united airframe.

The assembly scheme of the rib we use here is a sequential scheme, where each and every part of the rib structure is installed separately one by one to complete the whole structure.

RIB MOUNTING

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For the assembly of the given rib section of the aircraft, it is recommended to use the vertical jig. It is very easy to use with and we can able to work out from both sides of the jig.Before stepping to assembly process ensure that necessary procedures of manufacturing of all part is done and kept for assembling. Refer assembly diagram

For the assembly of the given rib section of the aircraft, it is recommended to use the vertical jig. It is very easy to use with and we can able to work out from both sides of the jig.

Before stepping to assembly process ensure that necessary procedures of manufacturing of all part is done and kept for assembling. Refer assembly diagram while assembling.

Prepare the jig

Install first the right cap (T-profile) using clamps and fix it rigidly.

Install the web of rib after installing right cap.

Then install the left cap (T-profile) and fix it firmly.

Install the caps using elements of special clamps.

Assembly holes are made to fix all the caps to the web of rib.

Now install the side cap of section M using special Holes.

The caps and web are clamped together tightly.

Now the two stiffeners are glued to one side of the web and other three stiffeners are glued to another side of the web.

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The special attachment plate is placed to the other side of the rib and it is glued initially and then it is clamped.

If necessary use guide holes at required places.

The attachment plates are fixed using special bolts

Then rivet the whole assembled structure using the rivet guns. Trailing edges of the rib is installed very carefully using assembly holes.

Avoid drilling after assembled because it would damage the whole structure.

Make sure that the contour shape of the web of the rib is not affected throughout the manufacturing and assembling process.

STAGES OF FORMATION OF RIB DIMENSIONS USING TEMPLATESThe scheme with application of mathematical model of a surface for manufacturing of rib using manual template method is shown below.

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THE SCHEME OF A RIB MANUFACTURING USING SURFACE MATHEMATICAL MODELScheme with application of mathematical model of a surface for manufacturing of rib using NPC method is shown below.

JIGA jig forms a guide for the tool used in machining operation. It is a type of tool used to control the location or motion of another tool. It is primarily used to provide repeatability, accuracy and interchangeability in the manufacturing of products. It does both functions that is holding the work and guiding a tool.

MAIN TYPES OF JIG

OPEN JIGIt is also called as plate jig, attaches to work piece without enclosing completely. It must be square or doughnut shaped and affixed over the work piece with clamps.DIAMETER JIGIt is used for cylindrical work pieces .It encloses a work piece in v-shaped groove.LEAF JIGS AND BOX JIGThese jigs completely enclose the work piece.Some other types of jig include,Trunion, Template, Guide, Sandwich, Angle plate, Universal

ASSEMBLY JIG

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A device or mechanism used in machine building for the attachment, securement, and correct alignment of parts and modules. In individual and small-lot production, general-purpose assembly jigs are used, including plates, assembly beams, knife-edged supports, try squares, clamps, and jacks. A set of such jigs serves as a basis for constructing stands used in assembling modules and complete machines. In mass and large-lot production, specialized assembly jigs are used for securing major components and modules (rotating and multi-position jigs) and for accurate and rapid assembly of parts and modules (single-position, multiposition, stationary, and mobile jigs). Assembly jigs are also used for the preliminary deformation of elastic elements, such as coil and leaf springs and split rings, and for making connections with interference fit. In progressive assembly, jigs are used for varying the positions of the objects being assembled. The use of assembly jigs improves the quality of the items produced, facilitates the work of assemblers, and increases productivity.

AIRCRAFT VERTICAL ASSEMBLY JIG DESIGN LAYOUT DIAGRAM

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THE DESIGNED VERTICAL FIN IN CATIA

Conclusion: The manufacturing and assembly process of vertical rib of the aircraft is briefed.

ECONOMICAL SECTION

1. ECONOMIC EFFICIENCY CHARACTERISTICS CALCULATIONThe economic calculations of the aircraft are done through required formulae or it can also be done on math cad which is pre-programmed already.

REQUIRED PARAMETERS:

Take-off mass – 39 .11 tonesEmpty mass = takeoff – fuel mass – payload mass = 26.98 tones

The items of expenses are:

Cost of materials and raw materials

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M o=1 . 95⋅104×md0.93×0.93 .32 log( N )

md-- Mass of the designed airplane in tonesN – Annual quantity of airplane

(N=8)md =39.11 tones

Thus M o=1 . 95⋅104×28 .650. 93×0 . 93 . 32 log(8)=391017.802834 $

Cost of devices onboard equipmentPurchased devices cost (PD) = 1.95(-1280+2.37 Vmax+14.15 Me.a) N-0.09

Me.a – mass of empty airplane in tonesMe.a=26.980tonsVmax–Maximum speed of airplane in km/hrVmax- 835 km/hrPD = 1651.575307$

Expenses for manufacturing special technological expensesGeneral expenses of labor manufacturing for special technological equipment

Tli = 0.87*1.03n*Me.a*106 hours n = total number of engines=2

= 24902081.34 dollars

Expenses in dollar Texp = Tli* C– price of one hour- (8)$

= 199216650.72 $

Expense for one airplane Toa = Texp/3(N1+N2)

N1 – production in the first year=3 N2 - production in the second year=5Toa = 11067591.706667

WagesExpenses on wages for work on direct manufacturing of airplane include a wage fund of industrial piece workers and workers creating the airplane by payment per hour.

W = 45200*Me.a0.903*M0.42*N-0.32*Kp

Me.a – mass of empty airplaneW =204074.65775$

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M - Maximum speed of airplane=0.7M , N– annual quantity of production Kp– coefficient which consider increase of labour charges of industrial workers during the production of airplane=0.64.

Indirect expenses Indirect shop expenses (IS)

IS = 5.837*W*N-0.129 = 839968.49613$

Indirect manufacturing expenses (IM) IM = 7.1*W*N-0.359=578606.52$

Taxes T = 0.375*0.7*W = 548063.227042

Total manufacturing expense Tme= Mo+PD+ToA+W+IS+IM+Tax= 13105937.063395$

Outside Manufacturing expense Ome= 5% (Tme)= 524237.482536$

Total Manufacturing cost price Tmcp= Tme+Ome=13630174.545931

Planned Profit15% Tmcp= 2044526.18189

Wholesale Price Tmcp+ profit= 15674700.727821

Value Added Tax 20% of wholesale price= 3134940.145564$

Total Wholesale price of enterprise with VAT Wholesale price + VAT= 18809640.873385$

Variable expenses:- Expenses which are changed (with manufacturing of airplanes) proportionally to the quantity of airplanes

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=Mo+PD+W+Tax= 650313.6333551$

Fixed expense :- expenses which are not changed with the manufacturing of airplane

= IS+IM+Toa+Om= 12979860.91238$

Analytical Calculation of break Even Point NBP=( Fixed expense)/(price – Variable expense)= 0.714777

The whole calculation is concluded in the table

S.No ArticlesExpenses

(in million US $)

1 Materials and Raw Materials0. 391017

2 Purchased devices , onboard equipments1651thousands

3 Manufacturing Repair of special technological equipment11.067591

4 Wage0. 204074

5 Indirect Shop expenses0. 839968

6 Indirect Manufacturing expenses0. 578606

7 Taxes0.548063

8 Total ( Manufacturing expenses)13.105937

9 Outside Manufacturing expenses0. 524237

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10 Total Manufacturing Cost Price13.630174

11 Profit (Planned) 2.044526

12 Wholesale Price15.67470

13 Value Added Tax3.13494

14 Wholesale Price with VAT18.809640

Conclusion:I have efficiently calculated the economic efficiency characteristics for my aircraft and found the cost of the aircraft and the amount of profit it that will be realized for the company annually.

SPECIAL ASSIGNMENT

INTERIOR CABIN LAYOUT AND SEATING ARRANGEMENTThe design process of the cabin is mainly characterized by the number of passenger’s accommodation and availability of volume, so it is necessary to find some of the base parameters for interior design of the cabinINITIAL DATA FOR CABIN DESIGN CONSIDERATION Number of passengers - 47 and can be increased up to 55 Payload mass which includes also mass of luggage - 4230 kg Diameter of fuselage – 2.5m

Factors influencing Fuselage seat dimensions are The seating layout or arrangement this can be selected based upon the prototype

statistical data which is briefed below, Cross section of the fuselage

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Diameter and length of the fuselageSEATING TYPE – 3 ABREAST 3 Abreast SEATING refers to the 2 × 1 seat arrangement, each row has two seats on right hand side and one seat to the left hand side of aircraft. This is selected based upon the prototype and diameter of the fuselage.A typical 3-abreast seating arrangement accommodates 24 to 45 passengers, but variant designs change that from 20 to 50passengers (e.g., ERJ145). Full standing headroom is possible; for smaller designs, a floorboard recess may be required. A floorboard recess could trip passengers when they are getting to their seat. Space below the floorboards is still not adequate for accommodating any type of payload. Generally, space for luggage in the fuselage is located in a separate compartment at the rear but in front of the aft pressure bulkhead (the luggage-compartment door is sealed).

FULFILLING REQUIREMENTS OF THREE ABREAST SEATING LAYOUT Full standing headroom is possible A floorboard recess may be required Space for luggage in the fuselage is located in a separate compartment at the rear but

before aft pressure bulk head Luggage compartment door is sealed if present Two cabin crews for 50 passengers

NUMBER OF ROWS

It is determined by the formula NO OF PASSENGENRS

ABREAST NUMBER =

473

= 15.66

15.66 can be rounded to either 16 , or

For 45 passengers = 453

= 15 rows

For two passengers = 22

= 1 row, it may be either 2 abreast or 3 abreast according to

designers choice.Therefore no of Rows = 16COMFORT LEVEL DETERMINATIONThe comfort level of the aircraft setting layout is the process of distributing the mass of the payload and its luggage capacity if the aircraft is designed with varying passengers numbers. In general there are two types of comfort level one is the low level and another is high level comfort.LOW LEVEL COMFORTIn low level comfort the number of passenger is increased by decreasing the luggage mass and compacting the cabin length. The low level comfort scheme of the designed aircraft is given below,

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The mass of luggage in low level comfort – 7 kg per passengerNumber of passengers – 55Mass of the single passenger – 70 kgTotal mass of passengers – 70 x 55 = 3850 kgTotal mass of luggage – 7 x 55 = 385 kgTotal Mass of payload = 3850 + 385 = 4230 kgHIGH LEVEL COMFORTIn high level comfort the number of passenger is decreased or set to the designed goal standard by increasing the luggage mass and providing more convenient seating arrangement inside the cabin. The high level comfort scheme of the designed aircraft is given below,The mass of luggage in high level comfort – 15 kg per passengerNumber of passenger – 47 The mass of single the passenger - 75 kgTotal mass of the passenger – 75 x 47 = 3525 kgMass of luggage – 15 x 47 = 705 kgHigh comfort level = Number of passengers x mass of the passenger + mass of luggage = 75 x 47 + 705 = 4230kgSTANDARD SEAT AND AISLE PITCH AND WIDTH

SEAT PITCH cm SEAT WIDTH cm AISLE WDTH cmECONOMY CLASS 71 to 81 46 to 51 43 to 41BUISNESS CLASS 84 to 91.5 53 to 56 56 to 63.5

The dimensions given above are general recommendation it varies according to operators and changes for different abreast seating of the aircraft.

DIMENSIONSSeat width, B (Left hand side) - 48.5 cmAisle width, A - 45.72 cmSeat width – B (Right Hand side) - 2 x 48.5 cmTotal elbow room - 20 cmGap between wall & seat, G - 5 cmCABIN DIMENSIONING FOR 3- ABREAST SEATINGThe cabin dimensioning depends on two main factors, Fuselage width Fuselage height Cargo compartment location

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So the selection of cabin width and floor height may vary according to the designed aircraft. But the normal cabin dimensioning for 3 abreast aircraft is given below we can use more or less value based on it and also we can refer our prototypes data for this selection.Total cabin width Wcabin - 215 cmTotal wall thickness T - 20 cmTotal fuselage width Wfuselage - 236 cmTypical fuselage height, Hfuselage – 215 cm

INTERIOR ARRANGEMENT – CROS SSECTION (TYPICAL)

The above picture depicts an example of 3 Abreast fuselage cross-section of Embraer 145 by comparing with these data’s my aircraft cabin is selected relative to the fuselage diameter, followed by the seat dimensions. Thus the whole aircraft seating layout with its cross section are drawn with all the necessary dimensions.DETERMINATION OF DESIGNED AIRCRAFT CABIN CROSS-SECTIONThe cabin cross-section of the designed aircraft is constructed based on the collected information and compared with our aircraft fuselage diameter, and then the cross section is drawn as given below,

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Total fuselage width – 2.5 mTotal fuselage width – 2.5 mSeat width – 0.45 mSeat pitch – 0.8277mCabin diameter – 2.27 mMaximum floor height – 1.85 mFloor width – 0.44 mGap between wall and seat - 0.09 mWall thickness = (fuselage diameter – cabin diameter) = 2.5-2.27 = 0.23m

DETERMINATION OF CABIN LENGTH FOR HIGH COMFORT LEVEL

The length of both the passenger and cockpit compartment is found statistically by selecting first selecting the cabin accessories for designed aircraft. The accessories placed inside aircraft varies on the right and left side so it is necessary to determine list of accessories and their quantities to place. Then the length of each accessory should be mentioned, along with the interior spacing and the lavatory dimension. The right hand side of the aircraft cabin includes the following,

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Closet - one closet = 0.80 mGalley – two galleys = 0.70 m each (2 X 0.70 = 1.4 m) Galley doors – one galley door (width) = 0.60 mSpace between the cockpit compartment and galley = 0.53mSpace between the second galley and first seat (RHS) = 0.30mTotal length of the occupied seat = Number of rows x seat pitch = 16 x 0.8293 = 13.21mSpace between last seat and lavatory = 0.30mLavatory length in side cross-section = 1mTOTAL LENGTH = 0.53 + 0.80+ 0.70+ 0.60 +0.70 +0.30 +13.21 + 0.30 + 1

Total length of the passenger cabin = 18.14m (RIGHT HAND SIDE)

The Left hand side of the aircraft cabin includes the following,

Passenger main door – 0.80 mSpace between cockpit compartment and main door – 0.46 mMain door length – 0.80 mSpace between main door and cabin attendant seat – 1.2879mSpace between cabin attendant seat to 1st row main seat – 0.82mCabin attendant seat length – 0.46Total length of the occupied seat = Number of rows x seat pitch = 16 x 0.8293 = 13.21mSpace between last seat and lavatory = 0.30mLavatory length in side cross-section = 1m

TOTAL LENGTH = 0.80 + 0.46 + 0.80 + 1.287 + 0.82 + 0.46 +13.21 + 0.30 + 1 = 18.14mTotal length of the passenger cabin = 18.14m (LEFT HAND SIDE)Therefore length on RHS = length on LHS

NOW adding the cockpit length which is equal to 1.90m, we getTotal length of cabin including cockpit = 18.14 + 1.90 = 20.04 mThen the final length is sketched for the foe calculated values by arranging the accessories in the cabin at appropriate places. The selection of those accessories may be from different manufacturers but it is compulsory to follow the rules and regulation of aviation federation. Some of the design parts used in this type of aircraft is given below , For the designed aircraft Galley no 1 is chosen as below

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Galley used in embraer 120For galley no 2:

Closet of the aircraft

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Service cart for providing food

Beverages cart:

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SEAT DESIGN

A two dimensional model of the chair is shown below for the reference to design a chair if needed but in practice the seats for an aircraft are selected based on the nominal values set previously so it is better to choose a good manufacturer for the selection of aircraft seat some examples are provided below

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SEAT SELECTION- ECONOMY RANGE SEATS

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CONCLUSION: Thus the cabin-cross section, length, seat dimensions are found and it is sketched briefly.

[194]

DIAGRAMS

LOAD CARRYING STRUCTURE

CENTRE OF GRAVITY

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GENERAL VIEW DIAGRAM

SEATING ARRANGEMENT

[196]

HYDRAULIC SYSTEM OF THE AIRCRAFT

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CONCLUSION OF THE PROJECT

Thus the selected aircraft is processed starting from the initial geometrical parameters following its aerodynamic characteristics, the loads acting on unit structure, integrated designing , manufacturing techniques, the hydraulic system, and even its economical valuation s are done each step by step according to the valuable lectures given by my professors and with the help of their textbooks. At last the designed aircrafts seating layout is considered and cabin length are found and sketched. The aircraft which is designed here follows all the rules and regulations of corresponding aviation federation with necessary limitations and has its own special requirements which are solely fulfilled according to the projects calculation and results.

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REFERENCES

1.DEVELOPMENT OF A PILOT PROJECT OF AN AIRCRAFT-Training guide BY A.K. Myalitsa, L.A. Malashenko, A.G. Grebenikov, E.T. Vasilevskiy, V.N. Klimenko, A.A. Serdyukov – Kharkiv: National Aerospace University Kharkov Aviation Institute

2. Aircraft Design - A Conceptual Approach 2nd ed. - D. Raymer (1992)

3. Torenbeek - Synthesis of Subsonic Airplane Design

4. Airplane design volume by Jhon Roskam

5. Aircraft Design Ajoy Kumar Kundu Queen’s University Belfast

6. Understanding aircraft structures By John Cutler- Jeremy Liber

7. Anderson Jr (2001) - Fundamentals of Aerodynamics

8. Mechnologies of aircraft manufacturing by Yu. M. Bukin, Vorobyov

9. http://triomatic.boico.com/Examples.htm.

10. www. Scribd.com