Konya Necmettin Erbakan University Faculty of Aviation and Space Sciences Aeronautical Engineering Design Conceptual Aircraft Design Konya 2018 1 Miraç POYRAZ 14030021035 Yusuf BAŞTOSUN 13030021005 Fatih YILMAZ 15030021038 2 ABSTRACT This report is prepared for aeronautical engineering design project in 2018-2019academic yearfall semester by Miraç POYRAZ (14030021035), Yusuf BAŞTOSUN (13030021005) and Fatih YILMAZ (15030021038). The project is about the conceptual design of a general aviation aircraft with twin engines. It is a known fact that aircraft design is the intellectual engineering process of creating a flying machine. Our designed aircraft must checks these steps: Certain specifications and requirements estabilished by potential users. Pioneer innovative ideas and technology. Design process starts when an airplane first materializes in peoples’ minds, ends when the finished product rolls out of manufacturer’s door. Three distinct phases of design are conceptual, preliminary and detail design. This project, as said upside, will contain only the conceptual design of an aircraft. If we want to explain conceptual design, we can describe it as this: “A set of specifications for concrete goal is given.” For more detail, we determine overal size, shape, weight and performance of the aircraft. Our top question on conceptual design is this: “Can the design meet the specifications?” and we will try to answer this question on this paper. For example; shape and location of the wings, vertical and horizontal tails, engine type, size and placement will be determined. 3 NOMENCLATURE AR b c c c.g. CD CD,i CD,0 Cfe CL Cl,max CL,max CM E e L/D L/D)max M n R R/C Re S Swet Sg Sref T/W TA TR V Aspect Ratio Span Specific Fuel Consumption Chord Length Center of Gravity Drag Coefficient Induced Drag Coefficient Parasite Drag Coefficient Skin Friction Coefficient Lift Coefficient Airfoil Maximum Lift Coefficient Wing Maximum Lift Coefficient Pitching Moment Coefficient Endurance Oswald Span Efficiency Lift to Drag Ratio Maximum Lift to Drag Ratio Mach Number Load Factor Range Rate of Climb Reynolds Number Wing Area Wetted Wing Area Ground Roll Distance Reference Area Thrust to Weight Ratio Thrust Available Thrust Required Velocity 4 NOMENCLATURE (Resuming) Vstall W/S W0 Wcrew Wfuel Wpayload x Stall Velocity Wing Loading Gross Weight Crew Weight Fuel Weight Payload Weight c.g Location (from nose of aircraft) NOMENCLATURE (Latters and Their Definitons) λ Λ α ρ ∞ c/4 1/h Taper Ratio Sweep Angle Angle of Attack Air Density Freestream Quarter Chord … per hour 5 Table of Contests Chapter I: Introduction 1.1 Purpose....................................................................................9 1.2 Requirements.......................................................................... 9 1.3 Mission Profile......................................................................... 10 1.4 Descriptıon of the Report........................................................ 11 Chapter II: Competitor Study 2.1 Introduction………………………………………………………………………... 11 2.2 The Research of Available Aircrafts......................................... 11 2.2.1 Citation XLS+ CE-560XL................................................11 2.2.2 Learjet 75 Model 45 ………………………………………………… 13 2.2.3 Legacy 500 EMB-550 ……………………………………………….. 15 2.2.4 Gulfstream 150 G150 ………………………………………………. 17 2.2.5 Falcon 2000S Falcon 2000EX …………………………………….19 2.3 Discussion and Conclusion ...................................................... 21 Chapter III: First Guess Sizing Study 3.1 Introduction ............................................................................22 3.2 Development ..........................................................................22 3.3 Discussion and Conclusion......................................................39 6 Chapter IV: Airfoil and Geometry Selection 4.1 Introduction...............................................................................40 4.2 Development............................................................................. 40 4.2.1 Airfoil Geometry ……………………………….......................... 40 4.2.2 Maximum Lift (Clean) ……………………………………………….. 41 4.2.3 Calculation of Design Lift Coefficient ………………………… 41 4.2.4 L/D And CL/CD ………………………………………………………….. 41 4.2.5 Thickness Ratio and Airfoil Types ………………………..……. 42 4.2.6 Airfoil Analysis and Selection ……………………………………. 44 4.3 Conclusion ……….........................................................................52 Chapter V: Thrust to Weight Ratio and Wing Loading 5.1 Introduction............................................................................... 54 5.2 Development .............................................................................54 5.3 Engine Selection……………….........................................................58 5.3.1 Williams FJ44-3A ………………………………..........................58 5.3.2 Williams FJ44-3A-24 …………………………………………………. 59 5.3.3 Garrett F109……………………………………………………………… 60 5.3.4 Pratt & Whitney Canada JT15……..…………………………….. 61 5.3.5 Williams FJ44-1A ………………………..…………..………………... 62 5.4 Engine Sizing ..............................................................................63 5.5 Discussion………………...................................................................63 Chapter VI: Initial Sizing 6.1 Introduction...............................................................................64 6.2 Refined Sizing ............................................................................64 7 6.3 Discussion……………….................................................................65 Chapter VII: Geometry Sizing Configuration 7.1 Introduction...............................................................................66 7.2 Development..............................................................................66 7.2.1 Tail Arrangement and Sizing…………………………………..67 7.2.1.1 Conventional Tail……………………………67 7.2.1.2 T-Tail………………………………………………67 7.2.1.3 Cruciform Tail…………………………………67 7.2.1.4 H-Tail………………………………………………67 7.2.1.5 V-Tail and Inverted V-Tail………….……68 7.2.1.6 Y-Tail……………………………………………...68 7.2.1.7 Twin Tails…………………………….…………68 7.2.2 Size of Horizontal and Vertical Tails………………………..71 7.2.1.1 Calculations of Horizontal Tail…..……71 7.2.1.2 Calculations of Vertical Tail ………...…71 7.3 Conclusion…………………………………………………………………………...71 Chapter VIII: Landing Gear Selections and Sizing 8.1 Introduction…………………………………………………………………………………73 8.2 Selection of Landing Gear…………………………………………………………….73 8.3 Calculation Criterias of Landing Gear……………………………………….....74 8.4 Tire Sizing…………………………………………………………………………………….75 8.5 Discussion………………………………………………………………..………………….77 Chapter IX: Model and Technical Drawing Pictures Of Our Business Jet.78 REFERENCES………………………………………………………………………………………..80 8 SECTION 1: INTRODUCTION 1.1 PURPOSE The target is to design a general aviation aircraft with twin engines, for customers’ requirements listed below. The design stages include airfoil parameters, flight mechanics etc. Procedure of design requires researches on these fields. All calculations were done to get the optimum conditions for the stage that is worked on. An aircraft should be designed in its best for all the aspects it contains. 1.2 REQUIREMENTS The requirements given by the customer are listed below. The design will be performed according to the requirements below. 1- Maximum lift to drag ratio will be taken equal to 14. 2- Cruise speed is equal to 850 km/h. 3- Range should be 3500 km. 4- Service ceiling should 45000 ft (13716 m) 5- 2 people will be contained as the crew member. 6- The aircraft will be selected as propeller driven A/C. 7- Rate of climb is 3500 ft/min (17.78 m/s). 8- Take-off distance won’t be more than 1000 m. 9-Historical trends (Wi / W(i-1)) will be taken equal to the values listed below: a)Engine start, Warm-up, Taxi-in and take off (W1/W0) = 0.970 b)Climb (W2/W1) = 0.985 c)Descent (W4/W3) = 0.995 d)Descent and landing (W6/W5)= 1.000 12- Specific fuel consumption will be taken from the table as pure turbojet, as c = 0.9 1/h 13- Maximum allowed passenger number is 8 with each having 20 kg luggages most. 9 1.3 MISSION PROFILE Since the aircraft will be designed as a propeller driven aircraft so the performance should be good enough to overcome the difficulties that each mission profile has. So, the performance parameters like weight, range, endurance etc. should be optimized for the conditions. The mission profile drawn below as: Figure 1.1. An example mission profile Here, the segments are: 0-1 : Take-off 1-2: Climb 2-3: Cruise 3-4: Combat 4-5: Descend (with loiter) 5-6: Ground attack 6-7: Climb 7-8: Cruise (at a higher altitude) 8-9: Descend (with loiter) 9-10: Landing But in this report, we will neglect combat, ground attack and randomized descend-cruises because of we are not working on fighter jet, so our mission profile will be looked as this: Figure 1.2. Our project’s mission profile 10 Our segments on Figure 1.2 can be named as: 0-1: Take-off 1-2: Climb 2-3: Cruise 3-4: Descend 4-5: Loiter (for 30 mins) 5-6: Descend and Landing 1.4 DESCRIPTION OF THE REPORT Aircraft design is a long process. To design an aircraft conceptually, one can follow steps in sequence strictly because the parameters are related to each other. For example, if the wing estimation is bad, we have bad results for thrust to weight ratio and wing loading. To get the logic during the whole design, the sequence below is followed, and detailed with sections: Section 1: Introduction Section 2: Competitor study Section 3: First guess sizing Section 4: Airfoil and geometry selection Section 5: Thrust to Weight Ratio and Wing Loading 11 SECTION 2: COMPETITOR STUDY 2.1 INTRODUCTION For any conceptual design paper, project’s writer team must examine the examples that manufactured before this one. Science is a fact that grows cumulatively, that shouldn’t be forget any time. Also in aviation, historical research has more necessity because the history of aviation was written with blood. So, the mistakes were made in the past shouldn’t be repeat. The close requirements to our aircraft were researched for different types of aircraft from the past. The reader will understand why these aircraft were selected among the millions of aircrafts. 2.2 THE RESEARCH OF AVAILABLE AIRCRAFTS 2.2.1 Citation XLS+ CE-560XL 2.2.1.1 General Explanation The Cessna Citation Excel (Model 560XL) is an midsize business jet built by Cessna, part of the Citation Family. Announced in October 1994. It first flew on February 29, 1996; certification was granted in April 1998 and over 900 have been delivered.With the success of Cessna's high-end Citation VII, the manufacturer saw a market for an aircraft with the Citation X's features but aimed at a more traditional market, where it would chiefly compete with twin-turboprop aircraft. The project was announced at the annual NBAA convention in October, 1994, and the prototype aircraft took off on its first flight on February 29, 1996. Federal Aviation Administration certification was granted in April 1998, by which time Cessna had over 200 orders for the aircraft. By the time the 100th Excel was delivered in August 2000, an aircraft was coming off the Wichita production line every three days. A total of 308 were built before production switched to the Citation XLS. The Citation XLS was the first "makeover" that the Excel received, with deliveries beginning in 2004. Besides a glass cockpit based on the Honeywell Primus 1000 EFIS avionics suite, the XLS featured the upgraded PW545B engines with increased performance. It was produced in 330 units. The Citation XLS+, or simply "Plus" configuration was another upgraded version of the aircraft which began delivery in 2008, with the inclusion of FADEC engine controls, improved PW545C engines, and a completely revised nose design similar to that found on the Citation Sovereign and Citation X. The Citation XLS+ features Collins Pro Line 21 Avionics and a four screen LCD EFIS display as opposed to the three tube (CRT) Honeywell display in the XL and the three screen LCD Primus 1000 in the XLS. 12 2.2.1.2 General Characteristics Crew:2 seats Capacity:9-12 people Length:16.0 m Wingspan: 17.17 m Height: 5.23 m Wing area: 34.5 m2 Airfoil (root): NACA 23014 mod Airfoil (tip):NACA 23012 Empty weight: 5086 kg Usefull Load: 4077 kg Max. takeoff weight: 9163 kg Powerplant: 2 × Pratt & Whitney Canada PW545C turbofans,4119 lbs (18.32 kN) 2.2.1.3 Performance Cruise speed: 816 - 856 km/h Range: 3541 km Service ceiling: 45000 ft (13716 m) Rate of climb: 3500 ft/min (17.78 m/s) Wing loading: 54.6 lb/ft2 Power/mass: 2.45 hp/lb 13 2.2.2. Learjet 75 Model 45 2.2.2.1 General Explanation The Learjet 45 (LJ45) aircraft is a mid-size business jet aircraft produced by the Learjet Division of Bombardier Aerospace. The Model 45 was the first all-new design since the original Learjet, and significantly altered the Learjet line. Through its four primary variants – the original Model 45, the Model 45XR, Model 40 and Model 40XR – it was the Learjet Division's principal product from the 1990s until the introduction of the Model 75 variant in 2012.The Model 45 was developed in the 1990s as a competitor in the "super-light" business jet category, a rival to the popular Cessna Citation Excel / XLS – but sacrificing the Cessna's stand-up room for the Learjet family's traditional high-speed performance. The Model 45 was Learjet's first completely new ("clean sheet") design since the company's first aircraft (all other models having been evolved from the original 1963 Learjet design, the Model 23). The Model 45 was developed to make Learjets fundamentally more competitive against newer designs from competing manufacturers. But, as a clean-sheet design (starting from scratch), being built to more rigorous (FAR Part 25) rules, than previous Learjets, the aircraft's development took substantially greater time than the development of previous Learjet models. The development of the LJ45 began in 1989, but was not announced by Bombardier until September, 1992. First flight of the prototype aircraft took place on October 7, 1995 – the 32nd anniversary of the first flight of the original Learjet 23. FAA certification was delayed, and finally granted in September 1997, with the first customer aircraft subsequently delivered in January, 1998.Flying Magazine reports that the Lear 45 was first certified under FAR Part 25 (transport category rules) in 1998. Initially, delays in production resulted in frustrated customers and lost or delayed revenues. Some customers' orders were delayed more than two years.By late 2006, Learjet had delivered over 300 Model 45s (40 of which were its improved version, the Model 45XR). By November 2010, over 500 of the Model 45 and 45XR, had been delivered. 14 2.2.2.2 General Characteristics Crew:2 seats Capacity:8-9 people Length:17.68 m Wingspan:14.58 m Height:4.30 m Wing area:28.95 m2 Airfoil: Learjet Airfoil (both tip and root) Empty weight: 6300 kg Max takeoff weight:9752 kg Powerplant:2 × Honeywell TFE731-20 turbofan, 3,650 lbs (15.57 kN) each 2.2.2.3 Performance Cruise Speed: 804 – 854 km/h Range: 3167 km Service Ceiling: 51000 ft (15545 m) Rate of Climb: 2800 ft/min (14.22 m/s) Wing Loading: 69 lb/ft2 Power/mass: 2.79 hp/lb 15 2.2.3. Legacy 500 EMB-550 2.2.3.1 General Explanation The Embraer Legacy 450/500 (EMB-545/EMB-550) are Brazilian mid-size business jets launched by Embraer in April 2008, the first of their size with a flat-floor stand-up cabin and fly-by-wire.At the August 2007 NBAA convention, Embraer unveiled a cabin mock-up of two concepts positioned between the $7 million Phenom 300 and the $26 million Legacy 600, called midsize jet (MSJ) and midlight jet (MLJ), positioned on 22% of the market in units. They should share their flat floor, stand-up cabin but the MSJ should be 5 feet longer to accommodate 8 passengers over a 2,800-nm range against 2,200-nm for the smaller version. The program was introduced in April 2008, Embraer planned to invest US$750 million and to introduce the larger model in 2012 and the smaller in 2013. Honeywell HTF7500E turbofans were selected along a Rockwell Collins Pro Line Fusion avionics suite integrated cockpit and a Parker Hannifin fly-by-wire flight control system. At the May 2008 European Business Aviation Convention and Exhibition, the larger was named Legacy 500 and priced at $18.4 million and the smaller Legacy 450, priced at $15.25 million. The variants have 95% systems commonality. An assembly line was officially opened in Melbourne, Florida on 2 June 2016, adding Embraer Legacy 450 and 500 production to the existing Phenom 100 and 300 line, along a completion center/flight-prep building. The first Legacy 450 on the line since May 16 should be delivered in mid-December. The facility will be able to assemble up to 96 Phenoms and 72 Legacys annually. The first Legacy 450 produced in Florida was delivered in December 2016; the fuselage is built in Botucatu in Brazil, and the wings in Évora, Portugal. The first Legacy 500 entered final assembly in January 2017 and was flown in July. Embraer will eventually move most of its Legacy 450/500 production in Florida but has not set up a schedule yet. At the August 2007 NBAA convention, Embraer unveiled a cabin mock-up of two concepts positioned between the $7 million Phenom 300 and the $26 million Legacy 600, called midsize jet (MSJ) and midlight jet (MLJ), positioned on 22% of the market in units. They should share their flat floor, stand-up cabin but the MSJ should be 5 feet longer to accommodate 8 passengers over a 2,800-nm range against 2,200-nm for the smaller version. The program was introduced in April 2008, Embraer planned to invest US$750 million and to introduce the larger model in 2012 and the smaller in 2013. Honeywell HTF7500E turbofans were selected along a Rockwell Collins Pro Line Fusion avionics suite integrated cockpit and a Parker Hannifin fly-by-wire flight control system. At the May 2008 European Business Aviation Convention and Exhibition, the larger was named Legacy 500 and priced at $18.4 million and the smaller Legacy 450, priced at $15.25 million. The variants have 95% systems commonality. 16 An assembly line was officially opened in Melbourne, Florida on 2 June 2016, adding Embraer Legacy 450 and 500 production to the existing Phenom 100 and 300 line, along a completion center/flight-prep building. The first Legacy 450 on the line since May 16 should be delivered in mid-December. The facility will be able to assemble up to 96 Phenoms and 72 Legacys annually. The first Legacy 450 produced in Florida was delivered in December 2016; the fuselage is built in Botucatu in Brazil, and the wings in Évora, Portugal. The first Legacy 500 entered final assembly in January 2017 and was flown in July. Embraer will eventually move most of its Legacy 450/500 production in Florida but has not set up a schedule yet. 2.2.3.2 General Characteristics Crew:2 seats Capacity:8-12 people Length:20.74 m Wingspan:20.25 m Height:6.44 m Wing area:40.21 m2 Airfoil (root): NACA 4412 Airfoil (tip): NACA 2412 Empty weight:10167 kg Max takeoff weight: 17399 kg Powerplant:2 x Honeywell HTF7500E turbofan, 6540 lbf (29.09 kN) each 2.2.3.3 Performance Cruise Speed: 863 km/h Range: 5788 km Service Ceiling: 45000 ft (13716 m) Rate of Climb: 3783 ft/min (19.22 m/s) Wing Loading: 88 lb/ft2 Power/mass: 2.69 hp/lb 17 2.2.4. Gulfstream 150 G150 2.2.4.1 General Explanation The wide-cabin high-speed Gulfstream G150 was designed, developed, and certified in a joint effort between Israel Aircraft Industries (IAI) and Gulfstream Aerospace. The Gulfstream G150 lineage began with the Astra 1125, Astra SP, Astra SPX (1986 – 2002) produced by IAI and then Gulfstream acquired the production line and rebranded the line to the Gulfstream G100. The Astra SPX/Gulfstream 100 had the greatest speed and longest range compared to its competitors but not the cabin comfort needed for a US transcontinental mid- cabin business jet. Accordingly, Gulfstream decided to enlarge the cabin in the G100 from 304 cubic feet to 521 cubic feet in the G150. As of September 2016, 36% of the Gulfstream G150 aircraft were purchased pre-owned by their current owners, the other 64% new. 5% are currently for sale with all of those (100%) under an exclusive broker agreement. When for sale, the average number of days on the market is 280. The Gulfstream G150 began production in 2005 and is still being produced presently. A total of 123 Gulfstream G150 aircraft have been manufactured with 119 of them in operation, one retired and 3 in various stages of the completion process including interior installation and exterior paint. Gulfstream Aerospace recently announced it has sold the last Gulfstream G150, marking the end of the mid-size aircraft’s more than 10-year production run. The final G150 will be delivered to a customer in mid-2017. There are 119 Gulfstream G150 aircraft in operation today. Most of these are owned outright (116) with 11.8% currently leased. North America has the largest percentage of Gulfstream G150 aircraft (83%), followed by Europe (6%). 18 2.2.4.2 General Characteristics Crew:2 seats Capacity:6-9 people Length:16.94 m Wingspan:16.64 m Height:5.54 m Wing area:29.41 m2 Airfoil (both root and tip):NACA 63A214 Empty weight:6442 kg Max takeoff weight: 11838 kg Powerplant:2 × Honeywell TFE731-40-R-200G turbofans, 18.9 kN (4,250 lbf) 2.2.4.3 Performance Cruise Speed: 896 km/h Range: 5462 km Service Ceiling: 45000 ft (13716 m) Rate of Climb: 3805 ft/min (19.33 m/s) Wing Loading: 82.3 lb/ft2 Power/mass: 2.95 hp/lb 19 2.2.5 Falcon 2000S Falcon 2000EX 2.2.5.1 General Explanation The Dassault Falcon 2000 is a French business jet and a member of Dassault Aviation's Falcon business jet line, and is a twin-engine, slightly smaller development of the Falcon 900 trijet, with transcontinental range.Variant which began testing in 2011 with short field characteristics. Landing distance has been reduced to 705 meters, opening up 50% more airports than other aircraft in this class. In May 2017, early 2013 models were valued $16-18 million and $26 million for late 2016 models while direct operating cost are $2,300-2,500/hr. 20 2.2.5.2 General Characteristics Crew:2 seats Capacity:8 - 10 people Length:20.23 m Wingspan:19.33 m Height:7.06 m Wing area:49 m2 Airfoil (both root and tip):NACA 64A204 Empty weight:9405 kg Max takeoff weight: 18597 kg Powerplant:2 × Pratt & Whitney Canada PW308C turbofans, 7000lbs (31.1 kN) each. 2.2.5.3 Performance Cruise Speed: 851 km/h Range: 6020 km Service Ceiling: 47000 ft (15500 m) Rate of Climb: 4535 ft/min (23.03 m/s) Wing Loading: 77.7 lb/ft2 Power/mass: 2.93 hp/lb 21 2.3 DISCUSSION AND CONCLUSION The characteristics that the researched 5 aircrafts are very clear by listing. The requirements are discussed in the previous part of the report. In conclusion, the importance of competitor study is to observe the characteristic experiences of the aircraft in use. Such an observation provides a variety of advantages in the design steps which are mainly iteration processes. It can be said that, nowadays, the design process is a process between re-engineering and discovery. • • • • • In our example, we fly in a medium-sized aircraft. But our flying range is below the majority of the airplanes. Our aircraft meets the standards of the speed segment. Service Ceiling is segment standards. Weight below segment standards 22 SECTION 3: FIRST GUESS SIZING 3.1 INTRODUCTION The calculations are done for range, endurance and weight are shown clearly in the next stage of this section. The calculations for different range, different loiter times, different amount of passengers, different take-off fuel consumptions and different climb fuel consumptions. The range values taken as 2975 km,3150 km, 3325 km, 3675 km, 3850 km and 4025 km.Loitering times are 27, 28.5, 31.5 and 33 minutes. Passenger amounts are 6-79-10 people. Take-off and climb fuel consumpitons will be considered from their values; ± %10 and ± %5. 3.2 DEVELOPMENT 3.2.1. Solution with given initial values W0 =Wcrew +Wpayload +Wpassenger +Wfuel +Wempty or ( Wcrew +Wpayload +Wpassenger ) W0 = Wfuel Wempty 1−( 𝑊0 )−( 𝑊0 ) 1- 2- Wcrew Wpayload Wpassenger W₃ W₂ = (2*80)* 9,81 = (10*20)* 9,81 = (8*80)* 9,81 = 1569,6 N = 1962 N = 6278,4 N −(𝟑𝟓𝟎𝟎𝒌𝒎)∗(𝟎.𝟗𝟏\𝒉) −𝑹∗𝑪 = e (𝟖𝟓𝟎𝒌𝒎\⼺∗(𝟏𝟐.𝟏𝟐𝟒) = 0.736 = e𝑽∗(𝑳\𝑫) (L\D) = 0.866*(L\D)max (L\D) = 0.866*14 = 12.124 −𝑬∗𝑪 3- W₄ W₃ = e(𝑳\𝑫)𝒎𝒂𝒙 4- Wfuel 𝑊0 = 1.06*(1-𝑊₀) 𝑊₆ 𝑊₀ = 𝑊₀ * =e −(𝟑𝟎𝒎)∗(𝟎.𝟎𝟏𝟑𝟏\𝒎) (𝟏𝟒) = 0.972 𝑊₆ 𝑊₁ 𝑊₂ 5₁ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 23 5- Statistical Rations 𝑊ᵢ 𝑊₍ᵢ⎽₁₎ Take Off Climb Cruise Loiter Discend Landing 0.97 0.985 W₃\W₂ W₄\W₃ 1 0.995 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.680) Wempty 𝑊0 = A * W₀C * Kvs = 0.680 = 0.339 Where A(for metric) = 1.4 C = -0.1 And Kvs =1 for fixed sweep wing, 1.04 for varible wing. As we choose fixed sweep wing, we used 1. 6- Wempty 𝑊0 = 1.4 * W₀-0.1 W₀ = (1−0.339−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; 1000 𝑘𝐷∗ 9,81 𝑚/𝑠^2 W0 Guess 60000 53000 52000 52065 52064,17 W0 Calculated 50287,4924 51832,7861 52080,28705 52063,96853 52064,1767 the result of W0 is 5307,26 kg = 11700.51 lbs. 24 3.2.2 Trade Analysis 3.2.2.1 Range Trade Analysis 1- For Range 2975 km W₃ W₂ ; −𝑹∗𝑪 −(𝟐𝟗𝟕𝟓 𝒌𝒎)∗(𝟎.𝟗 𝟏\𝒉) = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓𝟎 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ = 0,771 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₆ 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.759)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.712) W₀ = (1−0.305−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = 0.712 = 0.305 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 46000 45400 45428 45430,15 W0 Calculated 43598,72266 45304,93327 45436,83888 45430,62354 45430,14653 the result of W0 is 4631,0 kg = 10188,20 lb. 2- For Range 3150km W₃ W₂ ; −𝑹∗𝑪 −(𝟑𝟏𝟓𝟎 𝒌𝒎)∗(𝟎.𝟗 𝟏\𝒉) = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓𝟎 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) = 0,759 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.759)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.701) W₀ = (1−0.317−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = 0.701 = 0.317 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 25 W0 Guess 55000 48000 47540 47580 47586,2 W0 Calculated 46054,91272 47491,16848 47596,93344 47587,67311 47586,23885 the result of W0 is 4850,78 kg = 10671,72 lb. 3– For Range 3325 km W₃ W₂ ; −𝑹∗𝑪 −(𝟑𝟑𝟐𝟓 𝒌𝒎)∗(𝟓.𝟗 𝟏\𝒉) = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓𝟎 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) = 0,748 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.748)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.691) W₀ = (1−0.327−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = 0.691 = 0.327 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 50000 49500 49535 49530,9 W0 Calculated 48323,55396 49419,24549 49538,31435 49529,91779 49530,9009 the result of W0 is 5049,02 kg = 11107,85 lb. 26 4– For Range 3675 km W₃ W₂ ; −𝑹∗𝑪 −(𝟑𝟔𝟕𝟓 𝒌𝒎)∗(𝟎.𝟗 𝟏\𝒉) = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓𝟎 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) = 0,725 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.725)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.669) 𝑊₆ W₀ = (1−0.350−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = 0.669 = 0.350 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 54500 54600 54601,7 5– 𝑊₅ W0 Calculated 54498,00418 54628,32339 54602,10463 54601,65958 the result of W0 is 5565,9 kg = 12245,03 lb. For Range 3850 km ; W₃ W₂ −(𝟑𝟖𝟓𝟎 𝒌𝒎)∗(𝟎.𝟗 𝟏\𝒉) −𝑹∗𝑪 = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓𝟎 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) = 0,714 𝑊₆ 𝑊₀ 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ Wfuel 𝑊0 = 1.06*(1-0.658) W₀ = (1−0.362−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = (0.97)*(0.985)*(0.714)*(0.972)*(1)*(0.995) = 0.658 = 0.362 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 58000 57600 57640 57637,18 W0 Calculated 58390,56603 57537,97233 57647,4019 57636,4026 57637,17764 the result of W0 is 5875,35 kg = 12925,77 lb. 27 6– For Range 4025 km W₃ W₂ ; −𝑹∗𝑪 −(𝟒𝟎𝟐𝟓 𝒌𝒎)∗(𝟎.𝟗 𝟏\𝒉) 𝒌𝒎\𝒉)∗(𝟏𝟐.𝟏𝟐𝟒) = e𝑽∗(𝑳\𝑫) 𝑊₁ = e (𝟖𝟓 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ = 0,703 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₆ 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.703)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.648) W₀ = (1−0.373−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; = 0.648 = 0.373 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 60000 60750 60700 60698,85 W0 Calculated 62481,45303 60902,04713 60684,13024 60698,51749 60698,84863 Weight the result of W0 is 6187,44 kg = 13612,38 lb. Range Range 2975 3150 3325 3500 3675 3850 4025 Weight 45430,15 47586,2 49530,9 52064,17 54601,7 57637,18 60698,85 28 3.2.2.2 Loiter Trade Analysis 1- For Loiter 27 minutes ; −𝑬∗𝑪 −(𝟐𝟕 𝒎)∗(𝟎.𝟎𝟏𝟑 𝟏\𝒎) (𝟏𝟒) W₄ W₃ = e(𝑳\𝑫)𝒎𝒂𝒙 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.736)*(0.975)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.682) W₀ = (1−0.337−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ =e 𝑊₃ 𝑊₄ 𝑊₅ = 0.975 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.682 = 0.337 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 51600 51625 51625,7 W0 Calculated 50827,27953 51533,11893 51632,14862 51625,92337 51625,74913 the result of W0 is 5262,55 kg = 11577,63 lb. 2- For Loiter 28,5 minutes ; −𝑬∗𝑪 −(𝟐𝟖,𝟓 𝒎)∗(𝟎.𝟎𝟏𝟑 𝟏\𝒎) (𝟏𝟒) W₄ W₃ = e(𝑳\𝑫)𝒎𝒂𝒙 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.736)*(0.9739)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.681) W₀ = (1−0.3379−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ =e 𝑊₃ 𝑊₄ 𝑊₅ = 0.9739 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.681 = 0.3379 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 29 W0 Guess 55000 52000 51800 51820 51822,2 W0 Calculated 51065,40021 51777,91503 51827,7475 51822,75042 51822,20093 the result of W0 is 5282,59 kg = 11621,69 lb. 3- For Loiter 31,5 minutes ; −𝑬∗𝑪 −(𝟑𝟏,𝟓 𝒎)∗(𝟎.𝟎𝟏𝟑 𝟏\𝒎) (𝟏𝟒) W₄ W₃ = e(𝑳\𝑫)𝒎𝒂𝒙 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.736)*(0.971)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6789) W₀ = (1−0.34−1.4∗𝑊₀⁻⁰’ⁱ) then will be iterated as ; 𝑊₁ 〰₂ 5₁ =e 𝑊₃ 𝑊₄ 𝑊₅ = 0.971 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.6789 = 0.34 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 52250 52285 52285,95 W0 Calculated 51629,78768 52358,25166 52294,9967 52286,17973 52285,94054 the result of W0 is 5329,86 kg = 11725,69 lb. 30 4- For Loiter 33 minutes ; −𝑬∗𝑪 −(𝟑𝟑 𝒎)∗(𝟎.𝟎𝟏𝟑 𝟏\𝒎) (𝟏𝟒) W₄ W₃ = e(𝑳\𝑫)𝒎𝒂𝒙 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.985)*(0.736)*(0.9698)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.678) W₀ = (1−0.341−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ =e 𝑊₃ 𝑊₄ 𝑊₅ = 0.9698 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.678 = 0.341 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 52450 52515 52509,4 W0 Calculated 51902,95163 52639,19931 52524,4687 52508,02711 52509,44233 the result of W0 is 5352,64 kg = 11775,81 lb. Weight Loiter 27 28,5 30 31,5 33 Weight 51625,7 51822,2 52064,17 52285,95 52509,4 Loiter 31 3.2.2.3 Passenger Number Trade Analysis 1- For 6 Person W₀ ; 800 𝑘𝑔∗ 9,81 𝑚/𝑠^2 = (1−0.339−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; W0 Guess 45000 43600 43608 43607,3 43607,275 W0 Calculated 43244,45 43609,21 43607,07 43607,26 43607,275 the result of W0 is 4445,185 kg = 9799,955 lb. 2- For 7 Person W₀ ; 900 𝑘𝑔∗ 9,81 𝑚/𝑠^2 = (1−0.339−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; W0 Guess 45000 48000 47850 47870 47870,3 W0 Calculated 48650,01 47836,88 47875,56 47870,4 47870,3 the result of W0 is 4879,747 kg = 10758 lb. 3- For 9 Person W₀ ; 1100 𝑘𝑔∗ 9,81 𝑚/𝑠^2 = (1−0.339−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; W0 Guess 60000 56000 56190 56198 56197,8 W0 Calculated 55316,24 56246,28 56199,74 56197,79 56197,8 the result of W0 is 5728,627 kg =12629,46 lb. 32 4- For 10 Person ; W₀ 1200 𝑘𝑔∗ 9,81 𝑚/𝑠^2 = (1−0.339−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; W0 Guess 60000 60200 60275 60278 60278,24 W0 Calculated 55316,24 56246,28 56199,74 56197,79 56197,8 the result of W0 is 6144,584 kg =13546,48 lb. Weight Passenger Number 6 7 8 9 10 Weight 43607,275 47870,3 52064,17 56197,8 60278,24 Passenger Number 33 3.2.2.4 Take Off Fuel Consuption Trade Analysis 1- For + %10 ; 奸₆ W₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₁ 𝑊₀ = 1 - (0,03*1,1) = 0,967 𝑊₆ 𝑊₀ = (0.967)*(0.985)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.678) W₀ = (1−0.341−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ = 0.678 = 0.341 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 50000 53000 52350 52510 52509,4 W0 Calculated 53169,1 52386,37 52349,83 52509,29 52509,4 the result of W0 is 5352,644 kg = 11800,56 lb. 2- For -%10 ; 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₁ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ 𝑊₁ 𝑊₀ = 1 - (0,03*0,9) = 0,973 𝑊₆ 𝑊₀ = (0.973)*(0.985)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6822) W₀ = (1−0.3368−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; = 0.6822 = 0.3368 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 34 W0 Guess 50000 52000 51600 51595 51595,4 W0 Calculated 52000 51495,4 51594,3 51595,5 51595,4 the result of W0 is 5259,47 kg = 11595,14 lb. 3- For + %5 ; 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₁ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ 𝑊₁ 𝑊₀ = 1 - (0,03*1,05) 𝑊₆ 𝑊₀ = (0.9685)*(0.985)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.679) W₀ = (1−0.340−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; = 0,9685 = 0.679 = 0.340 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 52250 52285 52286 W0 Calculated 51629,78768 52358,25166 52294,9967 52286,17973 52285,92795 the result of W0 is 5329,86 kg = 11725,70 lb. 35 4- For - %5 ; 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₁ 𝑊₂ 𝑊₃ 𝑊₄ 𝑊₅ 𝑊₆ 𝑊₁ 𝑊₀ = 1 - (0,03*0,95) 𝑊₆ 𝑊₀ = (0.9715)*(0.985)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6811) W₀ = (1−0.3379−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; = 0,9715 = 0.6811 = 0.3379 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 51800 51824 51822,2 W0 Calculated 51065,40021 51777,91503 51827,7475 51821,75137 51822,20093 the result of W0 is 5282,27 kg = 11621,82 lb. Weight Take Off Fuel Consumption 0,967 0,9685 0,97 0,9715 0,973 Weight 52509,4 52286 52064,17 51838,7 51595,4 Take Off Fuel Consumption 36 3.2.2.5 Climb Fuel Consuption Trade Analysis 1- For + %10 ; 𝑊₂ 𝑊₁ = 1- (0.015*1.1) 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₁ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ 𝑊₆ 𝑊₀ = (0.97)*(0.9835)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6789) W₀ = (1−0.3403−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 𝑊₃ = 0.9835 𝑊₄ 𝑊₅ 𝑊₆ = 0.6789 = 0.3403 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 52350 52352 52352,8 W0 Calculated 51711,43448 52442,22056 52353,51575 52353,0116 52352,80995 the result of W0 is 5336,403 kg = 11740,34 lb. 2- For - %10 ; 𝑊₂ 𝑊₁ = 1- (0.015*0,9) 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.9865)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.678) W₀ = (1−0.3381−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ 𝑊₃ =0.9865 𝑊₄ 𝑊₅ 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.6810 = 0.3381 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 51850 51865 51866 W0 Calculated 51118,61922 51832,63033 51870,05487 51866,30462 51866,05467 37 the result of W0 is 5287,05 kg = 11631,52 lb. 3- For + %5 ; 𝑊₂ 𝑊₁ =1- (0.015*1,05) 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.98425)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6795) W₀ = (1−0.3397−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ 𝑊₃ 𝑊₄ = 0.98425 𝑊₅ 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.6795 = 0.3397 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 W0 Guess 55000 52000 52250 52220 52219,25 W0 Calculated 51548,3983 52274,55123 52211,49822 52219,03902 52219,22763 the result of W0 is 5323,06 kg = 11710,74 lb. 4- For -%5 ; 𝑊₂ 𝑊₁ = 1- (0.015*0,95) 𝑊₆ 𝑊₀ = 𝑊₀ * 𝑊₆ 𝑊₀ = (0.97)*(0.98575)*(0.736)*(0.972)*(1)*(0.995) Wfuel 𝑊0 = 1.06*(1-0.6805) W₀ = (1−0.3386−1.4∗𝑊₀⁻⁰’ⁱ)then will be iterated as ; 𝑊₁ 𝑊₂ 5₁ 𝑊₃ 𝑊₄ = 0.98575 𝑊₅ 𝑊₆ * 𝑊₂ * 𝑊₃ * 𝑊₄ * 𝑊₅ = 0.6805 = 0.3386 1000 𝑘𝑔∗ 9,81 𝑚/𝑠^2 38 W0 Guess 55000 52000 51975 51976 51975,94 W0 Calculated 51252,15334 51969,92584 51976,18429 51975,93386 51975,94889 Weight the result of W0 is 5298,26 kg = 11656,17 lb. Climb Fuel Consumption Weight 0.9835 0.98425 0,985 0.98575 0.9865 52352,8 52219,25 52064,17 51975,94 51866 Climb Fuel Consumption 39 3.4 DISCUSSION AND CONCLUSION The approaches done in development section give an estimation of the take off gross weight of the training aircraft that will be designed.If the requirements discussed in section 1 and the results got from the assumptions shown above are compared, that will look like taking Wo between 1150 and 1200 is a very good assumption. Because in this interval, the values provide the close values to requirements. Apart from the comparison, if we obsess the result we got from the previous part, it can easily be seen when payload weight increase, take off gross weight increase, too. In each research step, there is a good interval with respect to the very small error values shared. Also the relation between range and gross weight. There is a direct proportion between these two. As seen from the calculations, when the range increases, gross weight has a minor change. The next sections of the report will show that whether this gross weight gross weight assumption is a good assumption or not. 40 SECTION 4: AIRFOIL SELECTION 4.1 INTRODUCTION This section includes the calculations of Reynolds Number, cl-cd-alpha graphics, drag and lift coefficients therefore lift to drag ratios etc. to select the best airfoil for the aircraft with having the optimized parameters. 4.2 DEVELOPMENT 4.2.1. Airfoil Geometry Airfoil geometry can be characterized by the coordinates of the upper and lower surface. It is often summarized by a few parameters such as: maximum thickness, maximum camber, position of max thickness, position of max camber, and nose radius. One can generate a reasonable airfoil section given these parameters. Airfoil geometry is shown figure on the below. Figure 4.1. A typical airfoil shape 41 It should be considered to select fit airfoil for our subsonic aircraft. Our researches showed us we may change our airfoil during considerations in the next steps. In this section we consider coefficient of lift and t/c ratio to decide airfoil. The perfect AoA is 6 degree for cruise flight and also coefficient of lift should be around 0.5 due to the data got by the research. 4.2.2. Maximum Lift (Clean) The maximum lift coefficient of the wing will usually determine the wingarea. This in turn will have a great influence upon the cruise drag. Thisstrongly affects the aircraft takeoff weight to perform the design mission.Thus, the maximum lift coefficient is critical in determining the aircraftweight; yet the estimation of maximum lift is probably the least reliable ofall of the calculations used in aircraft conceptual design. Even refined windtunnel tests cannot predict maximum lift with great accuracy. Frequently anaircraft must be modified during flight test to achieve the estimated maximumlift. For high-aspect-ratio wings with moderate sweep and a large airfoil leadingedge radius, the maximum lift depends mostly upon the airfoil characteristics.The maximum lift coefficient of the "clean" wing (i.e., withoutthe use of flaps and other high-lift devices) will usually be about 900Jo of theairfoil's maximum lift as determined from the 2-D airfoil data at a similar Reynolds number. 4.2.3. Calculation of design lift coefficient The lift coefficient is a number that engineers use to model all the complex dependencies of shape, inclination, and some flow conditions on lift. This equation is simply a rearrangement of the lift equation where we solve for the lift coefficient in terms of the other variables. The lift coefficient Cl is equal to the lift L divided by the density ρ times half the velocity V squared times the wing area S. The lift coefficient Cl refers to the dynamic lift feature of a two-dimensional foil section, with the reference area’s place taken by the foil chord. 4.2.4. L/D And CL/CD Lift and drag are both aerodynamic forces, the ratio of lift to drag is an indication of the aerodynamic efficiency of the airplane. Aerodynamicists call the lift to drag ratio the L/D ratio, pronounced "L over D ratio." An airplane has a high L/D ratio if it produces a large amount of lift or a small amount of drag. Under cruise conditions lift is equal to weight. A high lift aircraft can carry a large payload. Under cruise conditions thrust is equal to drag. A low drag aircraft requires low thrust. Thrust is produced by burning a fuel and a low thrust aircraft requires small amounts of fuel be burned. As discussed on the maximum flight time page, low fuel usage allows an aircraft to stay aloft for a long time, and that means the aircraft can fly long range missions. So an aircraft with a high L/D ratio can carry a large payload, for a long time, over a long distance. For glider aircraft with no engines, a high L/D ratio again produces a long range aircraft by reducing the steady state glide angle at which the glider descends. 42 Figure 4.2. Wing Loading Ratio Values For Different Types of Aircrafts For cruise; W=L 𝑊=(𝜌∞×𝑉∞2×𝑆×𝐶𝑙)/2 Air density is 0.24 kg/m3 at 13716 m, and cruise speed is 850km/h(236.11 m/s). According to historical data, wing loading is 26lb/ft2(1244.884482 N/m2) for general aviation (single engine). Wing loading ratio for other types of aircrafts is shown in Figure 4.2. 𝑊 (ρ∞ ×V∞2 ×C𝑙 ) 𝑆 = 2 = 1244.884482 = (0.24×236.112 ×𝐶𝑙 ) 2 𝐶𝑙 =0.186 4.2.5. Thickness Ratio and Airfoil Types First we calculate mach number at 13716 m. Speed of sound is 294.9 m/sat 13716m. 236.11 𝑀= 294.9 = 0.8 The thickness ratio from historical data which is shown on the next page. 43 Figure 4.3. Thickness Ratio Change Due To Different Mach Numbers When we get useful thickness ratio from graphic,we can say that thickness ratio should be between 0.14 and 0.16. 44 4.2.6. Airfoil Analysis and Selection Last research was about historical aircrafts that was similar with designing aircraft. Airfoils which used in the past, analysed. Reynolds number should be estimated. Wingspan is estimated being between 15 and 18 and also aspect ratio is estimated between 6.5 and 8 from historical data soour chord length should be between 1,6 m and 2,1 m. Estimated Reynolds number should be between 6 million and 8 million. After now, we will use the given datas for our further applications: Chord Length 1.6m 1.82m 2.1m Reynolds Number 6*106 7*106 8*106 We will derive these chord lengths and Reynolds numbers with the airfoils we choose, NACA 2412, NACA 4412 and NACA 23012. After some applications on Xfoil program, we gather these datas and tables: Reynolds Number 6*e6 7*e6 8*e6 Clmax 1,8825 1,9029 1,9204 Cdmin 0,00519 0,00514 0,00516 α (Cdmin) 1 2 1 α (Cl max ) 19 19 20 (Cl/Cd)max 126,7247 129,5386 132,8453 NACA 2412 datas 45 Cl / α 2 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 -10 -5 0 5 10 Re : 6*e6 15 Re : 7*e6 20 25 30 35 25 30 35 Re : 8*e6 Cd / α 0,45 0,4 0,35 0,3 0,25 0,2 0,15 0,1 0,05 0 -10 -5 0 5 Re : 6*e6 10 15 Re : 7*e6 20 Re : 8*e6 46 Cl/Cd α 180 160 140 120 100 80 60 40 20 0 -10 -5 0 5 10 Re : 6*e6 15 Re : 7*e6 20 25 30 35 25 30 35 Re : 8*e6 Cm / a 0 -10 -5 0 5 10 15 20 -0,05 -0,1 -0,15 -0,2 -0,25 -0,3 Re : 6*e6 Re : 7*e6 Re : 8*e6 47 Reynold Number 6*e6 7*e6 8*e6 Clmax 1,9372 1,9568 1,9728 Cdmin α (Cdmin) 0,00523 3 0,00527 3 0,00538 2 NACA 4412 datas α (Cl max ) 19 19 19 (Cl/Cd)max 157,4952 156,5612 169,4373 48 Cl / a 2 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 -10 -5 0 5 10 Re : 6*e6 15 Re : 7*e6 20 25 30 35 25 30 35 Re : 8*e6 Cd / a 0,45 0,4 0,35 0,3 0,25 0,2 0,15 0,1 0,05 0 -10 -5 0 5 Re : 6*e6 10 15 Re : 7*e6 20 Re : 8*e6 49 Cl/Cd - a 180 160 140 120 100 80 60 40 20 0 -10 -5 0 5 10 Re : 6*e6 15 Re : 7*e6 20 25 30 35 25 30 35 Re : 8*e6 Cm / a 0 -10 -5 0 5 10 15 20 -0,05 -0,1 -0,15 -0,2 -0,25 -0,3 Re : 6*e6 Re : 7*e6 Re : 8*e6 50 Reynold Number 6*e6 7*e6 8*e6 Clmax 1,8321 1,8597 1,8793 Cdmin α (Cdmin) 0,00533 3 0,00524 3 0,00518 3 NACA 23012 datas α (Cl max ) 19 19 19 (Cl/Cd)max 147,3503 149,4293 152,6409 51 Cl / a 2 1,8 1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 0 -5 0 5 10 15 Re : 6*e6 20 Re : 7*e6 25 30 35 40 30 35 40 Re : 8*e6 Cd / a 0,45 0,4 0,35 0,3 0,25 0,2 0,15 0,1 0,05 0 -5 0 5 10 Re : 6*e6 15 20 Re : 7*e6 25 Re : 8*e6 52 Cl/Cd - a 180 160 140 120 100 80 60 40 20 0 -5 0 5 10 15 Re : 6*e6 20 Re : 7*e6 25 30 35 40 30 35 40 Re : 8*e6 Cm / a 0,05 0 -5 0 5 10 15 20 25 -0,05 -0,1 -0,15 -0,2 -0,25 -0,3 Re : 6*e6 Re : 7*e6 Re : 8*e6 53 4.3. CONCLUSION The geometric characteristic of airfoil is listed below: b=17.79m c=2.1m AR=9.043 Re=8000000 S=36.5 m2 As a result of our analysis, NACA 4412 was selected as our airfoil. The factors of selecting this airfoil were Cl/Cd values relations and Cl-α graphic interpretation. Because the specified airfoils have similar values of stall speed and stall angle.In this case, we needed to look at the values of Cl/Cdaccording to the different airfoils. We can list this values in the table. AIRFOIL TYPE NACA 4412 NACA 2412 NACA 23012 Cl/Cd VALUE 169.4373 132.8453 152.6409 Table 4.1. The Comparison Of Airfoils Due To Cl/Cd Values Other specifications may reveal as: • • • • • • Naca 23012 is unstable as Cm passes to positive side The Cl / Cd ratio of Naca 4412 is the best. Cl is the best value of Naca 4412. The angles of the Naca 4412 are suitable for our aircraft. Chord length maximum Cl / Cd due to the maximum 2.1 was seen suitable. Chord length is maximum 2.1 due to maximum Cl. 54 SECTION 5: PERFORMANCE PARAMETERS CALCULATIONS AND ENGINE SELECTION 5.1 INTRODUCTION In this chapter, wing loading and thrust to weight ratio will be estimated by using variable parameters. Following different parameters as condition of cruise (L=W) and landing distance. Than the minimum wing loading will be selected. This yields to the selection of the engine and area of wing. At last, summary of the critical performance parameters will be listed. 5.2 DEVELOPMENT 5.2.1 Wing Loading Calculated by Cruise Parameters In most airplane designs, wing loading is determined by consideraions of V stall and landing distance. However, W/S also plays a role in the maximum velocity of the airplane. Vmax increases as W/S increases. For our current airplane design, which is a businessjet aircraft, the primary constraints on W/S will be Vstall, rate of climb and takeoff distance. Assume that Vstall = 150 km/h(136.7 ft/sec) from the information gathered from historical datas. 2𝑊 Vstall = √(ρ 1 ) 𝑆 CLmax CL,max is found previous chapter as 1,9728 in 2D. For AR>5 CL,max= 0.9* Cl,max CL,maxis found 1.776 for this chapter. 𝑊 𝑆 𝑊 𝑆 1 = 2 * ρ * Vstall2 * CLmax = (0,5)*(0,002377)*(136.7)2*(1,776) = 39.444lb/ft2 5.2.2 THRUST TO WEIGHT RATIO The value of T/W determine in part the take-off distance rate of climb and stall velocity. First our take-off distance which is specified as 3280 ft to clear a 35 ft obstacle estimate the ground roll. 1,21∗(𝑊⁄ ) 𝑆 sg = 𝑔∗ρ∞∗CL𝑚𝑎𝑥∗( 𝑇⁄ 𝑊) (1,21)∗(39.444) sg = (32,2)∗(0,002377)∗(1.776)∗(𝑇⁄ 𝑊) 351.11 = (𝑇⁄𝑊) 55 T varies with velocity as it does for a jet driven airplane, the value of T/W is assumed to be that for a velocity V∞ =0.7*VLOis the lift of velocity, taken as VLO=1.1*Vstall. Vstall must be calculated again using the take-off CLmax 2𝑊 Vstall = √(ρ 𝑆 1 39.444∗2 ) = √(0,002377)∗(1,776) = 136.7 ft/sec CLmax The flight path radius is calculated as the following; R= 6,96∗(𝑉𝑠𝑡𝑎𝑙𝑙)2 𝑔 = 6,96∗(136.7)2 32,2 = 4039.153 ft The included flight path angle is calculated as the following; ѲOB = cos-1(1 – ℎ𝑂𝐵 𝑅 ) Where hOB is the obstacle height hOB=35 ft so 35 ѲOB = cos-1(1 – 2805.132) = 7.55ᵒ The airborne distance is calculated as the following; sa = R*(sinѲOB) = (4039.153)*(sin(7.55ᵒ)) = 530.58 ft So we have; 243.83 sg + sa = 3280 = (𝑇⁄𝑊) + 441.72 𝑇 351.11 (𝑊)0,7VLO = 3280−530.58 = 0,127 This is the value of required T/W at a velocity V∞ = 0,7*VLO = 0,7*(1,1)*Vstall V∞ = (0,7)*(1,1)*(136.7) = 105.26 ft/sec At this velocity, the power required to take off at the gross weight W0=11700.51 lb. 𝑇 PR = T* V∞ = 𝑊 * W0 * V∞ = (0.127)*(11700.51)*(105.26) = 156411.17 lb.ft/sec P𝑅 T = V∞ = 156411.17 105.26 = 1486 lb = 674 kg*9,81 m/s2 = (6612.1 N) Next, let us consider the constraint due to the specified rate of climb of 17,78 m/s at sea level. Here, we need to make an estimate of the zero-lift drag coefficient, CD,0. From Figure 5.1 for twin engine general aviation airplanes the ratio of wetted area to the wing reference area is approximately Swet/Sref= 5.0-7.0 we choose 5.8. 56 Figure 5.1:Ratio of wetted surface area to reference area for a number of different airplane configurations. Equivalent skin friction drag for a variety of aircrafts The skin-friction coefficient C fe is shown as a function of Reynolds number in Fig. Reynolds number for us is 106. For this case Cf,e = 0.005. CD,0 = S𝑤𝑒𝑡 S * Cf,e = (5.8)*(0,005) = 0,029 We also need on estimate for the coefficient K that apperass in the drag polar 57 CD = CD,0 + K*CL2 𝐶 2 𝐿 K = k1+k2+k3 = k1+k2+𝜋𝑒𝐴𝑅 Let us estimate the value of K to be consistent with the earlier assumed value of (L/ D)max = 14. 1 (DL )max = 14 = √4∗Cᴅ,0∗𝐾 We have K= 1 1 4*𝐶𝐷,0 ∗(𝐿 ⁄𝐷 )𝑚𝑎𝑥 2 = 4*(0,029)∗142 = 0,044 1 AR = 𝜋∗0,8∗0,0773 = 9.043 Finally to return to the consideration of rate of climb. Desired rate of climb is 17,78 m/s (58 ft/sec) from our design requirements. (𝑅 ⁄𝐶 )max = μ∗P W = 58 + μ∗P W 2 𝐾 – [ρ √3𝐶 ∞ 𝑊 𝐷,0 𝑆 1/2 ] 1,155 *(L⁄D)max 2 0,044 1,155 [(0,002377) ∗ √3*0.029 ∗ (39.444)]1/2* 14 = 70.67 ft/sec P= W∗70.67 μ = (11700.51)∗70.67 1 = 826875 lb.ft/sec Since 550 lb.ft/s = 1 hp P= 826875 550 = 1503.41 hp = 1121.092 kW To satisfy the constraint on rate of climb, the power must be P ≥ 1458.522 hp T ≥ 10.65 kN 58 5.3 ENGINE SELECTION 5.3.1 Williams FJ44-3A Thrust: 2820 lbf = 12544 N Dry Weight: 535 lbs = 243 kg Overall Length: 62.4 in. = 1584.96 mm Approximate Fan Diameter: 22.9 in = 582 mm 59 5.3.2 Williams FJ44-3A-24 Thrust: 2490 lbf = 11076 N Dry Weight: 535 lbs = 243 kg Overall Length: 62.4 in. = 1585 mm Approximate Fan Diameter: 22.9 in. = 582 mm 60 5.3.3 Garrett F109 Maximum thrust: 1330 lbf = 5920 N Overall Length: 942.34 mm = 37.1 in Approximate Fan Diameter: 523 mm = 20.6 in Dry weight: 182 kg = 400 lbs Specific fuel consumption: (At max takeoff) 11.10 mg/Ns (0.392 lbs/hr.lbf) 61 5.3.4 Pratt & Whitney Canada JT15D Maximum thrust: 12920 N = 2904.532 lbf Length: 60.5 inches = 1600mm Diameter: 27 inches = 520mm Dry weight: 630 pounds = 287kg Specific fuel consumption: 0.562 lb/(lbf·h) at max, 0.552 lb/(lbf·h) at cruise (typ) 62 5.3.5 Williams FJ44-1A Maximum thrust: 8452 N = 1900 lbf Length: 53.3 inches = 1354 mm Diameter: 20.9 inches = 531 mm Dry weight: 460 pounds = 209 kg Specific fuel consumption: 0.456 lb/(lbf·h) at max, 0.552 lb/(lbf·h) at cruise (typ) 63 5.4 ENGINE SIZING We choose Williams FJ44-1A for the reasons that we talked about in 5.5 Conclusion section. For having a safely protected engine-box, we need to multiply by 1.1 of engine’s all measurements. Therefore: Length: 53.3 inches = 1354 mm Engine box Length = 1.1*1354 = 1489.4 mm Diameter: 20.9 inches = 531 mm Diameter Length = 1.1*531 = 584.1 mm 5.5 CONCLUSION Paremeters of engine selection is High available thrust Low fuel consumption Low weight to get most available engine. If considered about all these paramaters, best selection becomes Williams FJ44-1A because of having an okay thrust and also the lowest weight among options. 64 SECTION 6: INITIAL SIZING 6.1 INTRODUCTION In this small part, we are going to discuss about empty weight fraction. We calculated this one right back on section 2. But after that, we added an engine to our aircraft and decided the airfoil we are going to use. So, we need to approve that our empty weight fraction is still gives what we want. 6.2 REFINED SIZING EQUATION Empty weight is again expressed as an empty weight fraction but fuel weight is calculated directly. 𝑊0 = 𝑊𝐶𝑟𝑒𝑤 + 𝑊𝑃𝑎𝑦𝑙𝑜𝑎𝑑 + 𝑊𝐹𝑢𝑒𝑙 + 𝑊𝐸𝑚𝑝𝑡𝑦 𝑊𝐸𝑚𝑝𝑡𝑦 𝑊0 𝑊0 𝑊𝐸𝑚𝑝𝑡𝑦 We know every part of our equation except 𝑊 part. We are going to get a help for it, by 𝑊0 = 𝑊𝐶𝑟𝑒𝑤 + 𝑊𝑃𝑎𝑦𝑙𝑜𝑎𝑑 + 𝑊𝐹𝑢𝑒𝑙 + 0 using a table that took place on our lecture slides. We are going to have the jet transport coefficient on this one. You can find the calculation step by step on the next page. 65 𝑊𝑒 𝑇 𝐶3 𝑊0 𝐶4 𝐶1 𝐶2 = (𝑎 + 𝑏𝑊0 ∗ 𝐴𝑅 ∗ ( ) ∗ ( ) ∗ 𝑀𝑚𝑎𝑥 𝐶5 ) 𝐾𝑣𝑠 𝑊0 𝑊0 𝑆 𝑊𝑒 = (0.32 + 0.66 ∗ 11700.51(−0.13) ∗ 9.043(0.3) ∗ 0,127(0.06) ∗ 39.444 (−0.05) 𝑊0 ∗ 0.80.05 )1.00 𝑊𝑒 = 0.58 𝑊0 6.3 CONCLUSION 𝑊 After the Chapter 2’s weight estimation, we found 𝑊𝑒 as 0.56. The difference is %3.57, so we 0 agreed not to iterate one more. We decided we are okay to go. So, our engine is still gives what we wanted from it. 66 SECTION 7: GEOMETRY SIZING CONFIGURATION 7.1 INTRODUCTION After selecting both airfoil and the engine, now it is time to size and configure the tail surfaces. The horizontal and vertical configuration will be discussed in this section. This section is related to fuselage too but the others will be mentioned in the next section. The platform areas, aspect ratios, spans, chords and taper ratios of the horizontal and vertical tails will be calculated. 7.2 DEVELOPMENT We are going to calculate the crucial parts of the aircraft which are tail configurations and fuselage length. Starting geometry sizing with finding length of fuselage. As seen the below figure, the length of fuselage can be calculated. Designing aircraft length must be found with using general aviation – twin engine. 𝑙𝑓 = 𝑎𝑊0 𝑐 𝑙𝑓 = 0.86 ∗ 11700.51 0.42 𝑙𝑓 = 43.7 𝑓𝑡 = 13.4 𝑚 Fuselage length is found 13.4 m from equation. Now finding the size of tail sections with using length of fuselage, aspect ratio and surface area of wing. You can find the related calculations on the other page. 67 7.2.1 TAIL ARRANGEMENT AND SIZING You can find all the tail types in the figure upside. We are going to talk about them below. 7.2.1.1 Conventional Tail Provides adequate stability and control at low weight. However, the horizontal tail must be positioned behind the vertical so that its wake does not mask the rudder at high α, important for spin recovery. As a rule of thumb, 1/3 of the rudder should be out of the wake. 7.2.1.2 T-Tail Heavier than a conventional tail because the vertical tail must be strengthened to support the horizontal tail. Due to endplate effect, induced drag is less and the vertical tail can be sized smaller. 7.2.1.3 Cruciform Tail Cruciform tail is a compromise between a conventional tail and a T-tail. It avoids proximity to jet exhausts or exposes the lower part of the rudder to undisturbed air during high α flight or spins. 7.2.1.4 H-Tail An H-tail is used to position vertical tails in undisturbed air during high α flight or to position the rudders in the propwash in order to increase their effectiveness. It is heavier than a conventional tail but the endplate effect results in a lighter horizontal tail. 68 7.2.1.5 V-Tail and Inverted V-Tail In a V-tail, the wetted area is reduced. In theory, horizontal and vertical tail surfaces are found from the Pythogoran theorem. Research shows that the total wetted area of a V-tail is the same as that for separate horizontal and vertical tails. The resulting force pushes the tail to the left and the nose to the right as desired. However, the same force produces a roll moment towards the left adverse roll-yaw coupling. In an inverted V-tail, the adverse roll-yaw coupling problem is solved. 7.2.1.6 Y-Tail In a Y-tail, the additional surface contains the rudder, while V surfaces provide only pitch control. Avoids the complexity of the ruddervators and reduces the interference drag compared to the conventional tail. Twin tails position the rudders away from the aircraft centerline avoiding being blanketed by the wing or the fuselage at high α. 7.2.1.7 Twin Tails Twin tails reduce the height of the vertical tail. Twin tails are usually heavier than conventional tails but are often more effective. Other configurations like control-canard, lifting canard, tandem wing, flying wing are also possible. 69 For our tail type, we thought its better to choose conventional tail. It’s more simple than the others, might we have a little less efficiency from our tail but it is still the best of our options because of its weight, cost and easy usage. Now, we are ready to calculate our horizontal/vertical tail volume ratios. You can find their equations down below. CHT = CVT = 𝑙𝐻𝑇 ∗S𝐻𝑇 𝑐̅𝑆 𝑙𝐻𝑇 ∗S𝐻𝑇 bS For calculating SVT and SHT: SHT = C𝐻𝑇∗c∗S L𝐻𝑇 and 𝑆𝑉𝑇 = C𝑉𝑇∗b∗S L𝑉𝑇 For determining our CHT and CVT values, we are going to use the table that is below. Our jet is carrying the properties of general aviation-twin engine. CHT will be equal to 0.8 and CVT will be equal to 0.07. Now we can calculate our remaining calculations. (LHT and LVT is equal to tail arm and it is the half of the fuselage length.) SHT = SVT = C𝐻𝑇∗c∗S L𝐻𝑇 C𝑉𝑇∗b∗S L𝑉𝑇 = = 0,80∗2.1∗36.5 6,7 = 9.152 m2 0,07∗17.79∗36.5 6,7 = 6.78 m2 70 7.2.2 Size of Horizontal and Vertical Tails For our sizing parameters on our horizontal and vertical, we are going to choose our values from “Others” part. Therefore our values will be: Horizontal Tail Values Aspect Ratio 4 Vertical Tail Values Taper Ratio 0.5 Aspect Ratio 1.6 Taper Ratio 0.5 7.2.2.1 Calculations of Horizontal Tail b2 AR = S [ 𝐻𝑇 𝐻𝑇 4 = 9.152 => bHT = 6.05 (𝐶𝑟 𝑡 + 0.5𝐶𝑟 𝑡) 2 ] x b = SHT Crt = 2.017 m b 1+2𝜆 γHT = 6* 1+𝜆 = 2 CHT = 3*Crt[ 2 b => ; [ (𝐶𝑟 𝑡 + 0.5𝐶𝑟 𝑡) 2 ] x (6.05) = 9.152 Ctt = 4.034 m 17.79 1+2∗0,5 6 => * 1+0,5 (1+λ+λ2 ) ]» (1+λ) » γHT = 3.953 m (1+0,5+0,52 ) 2 CHT = 3*2.017[ (1+0,5) ]= 1.57 m 7.2.2.2 Calculations of Vertical Tail ARVT = H𝑉𝑇 2 => HVT2 = (1,6)*(6.78) => HVT = 3.294 m S𝑉𝑇 2∗S𝑉𝑇 2∗6.78 Crvt = (1+λ)H𝑉𝑇 = (1+0,5)3.294 = 2.744 m Ctvt = 0,5Crvt = 0,5*2,27 = 1,135 m The vertical location of the mean aerodynamic chord of the vertical tail, referenced to the root chord ZVT = 2H𝑉𝑇 (1+2λ) 6 => (1+λ) ZVT = 2∗3.294 (1+2∗0,5) 6 (1+0,5) = 1.464 m The mean aerodynamic chord for the vertical tail is 2 (1+λ+λ^2) CVT = 3*Crvt[ (1+λ) ] => 2 (1+0,5+0,52 ) CVT = 3*2,27[ (1+0,5) ] = 1.766 m 71 7.3 DISCUSSION AND CONCLUSIONS In this study the tail configuration was selected and sized. For the conceptual design case, as here, the best approach is the above applications supported by the historical data reached from references of tail sizing subject. The detailed stability and control analyses can only be done at the detail design sections of the aircraft design process. 72 SECTION 8: LANDING GEAR SELECTION AND SIZING 8.1 DISCUSSION AND CONCLUSIONS In this section, the reader will see how the calculation are done for landing gear location, weight estimation to compare whether the estimation of section 1 is good or not and more importantly centre of gravity location. These calculations are so important for conseptual design at all. With these calculations and results taken, it will be learned that where to place the components discussed on the aircraft. 8.2 SELECTION OF LANDING GEAR − Cabin floor is horizontal when the airplane is on the ground. − Forward visibility is improved for the pilot when the airplane is on the gound. − The CG is ahead of the main wheels and this enhances stability during the ground roll. Because of these reasons, we made our choose on tri-cycle landing gear. 73 8.3 CALCULATION CRITERIAS OF LANDING GEAR • The length of the landing gear must be set so that the tail doesn’t hit the ground during landing. • This is measured from the wheel in the static position assuming an angle of attack for landing that gives 90% of maximum lift, usually 10° -15°. • The tipback angle is the maximum aircraft nose-up attitude when the tail is touching the ground and the landing gear strut is fully extended. • To prevent the aircraft from tipping back on its tail, the angle of the vertical from the main wheel position to the cg should be greater than the tipback angle or 15°, whichever is greater. • Tipback angle should not be greater than 25°, otherwise porpoising will occur and a high elevator deflection will be required for rotation during takeoff. • This means that more than 20% of aircraft weight is carried by the nose wheel. • The optimum range for the percentage of aircraft weight that is carried by the nose wheel is 8-15% for the most aft and most forward CG positions. • If it is less than 5% there won’t be enough traction to steer. Overturn angle is a measure of the aircraft’s tendency to overturn when turned around a sharp corner. • This is the angle between the CG and the main wheel seen from the rear. This angle should not be greater than 63°. 74 8.4 TIRE SIZING • Tyre size depends on the load carried by each tyre. With the help of SolidWorks software, we made the centre of gravity locations visible and by that, we found out FWD c.g (Forward Centre of Gravity) and AFT c.g. (Aircraft Centre of Gravity) values. We choose the value of b as 4; so after all of these acceptions we found out our values as this: Na = 3.2 m Mf = 1.2 m Ma = 0.8 m With the reveal of Na, Mf and Ma values, we found out our maximum static load on main wheels. • Maximum static load (main) = W 𝑁𝑎 𝐵 3,2 = 5307* 4 = 4246 kg After calculated this static load value, we are going to use the table below to found out our diamater and width of landing gear. 75 Diameter for main wheel: A*WwB = 8.3*1061,50.251 = 47,70 cm (14,55 in.) Width for main wheel A*WwB = 3.5*1061,50.216 = 15,76 cm (4,86 in.) • Nose tyres are 60-100% the size of the main tyres. Diameter for nose wheel = 0.8*Main teker diameter = 0.8*14,55 = 11,64 in. Width for nose wheel = 0.8*Main teker width = 0.8*4,86 = 3,88 in. With the sizing values of our tires, now we can look the below table for choosing our main tires: According to the table; nose tire selected as Type 3 (5.00-4), main wheels selected as Type 7 (16*4.4). 76 8.5 DISCUSSION With the selection of landing gear, our aircraft is came together as one piece. At the next section, we are going to put the pictures of our aircraft’s drawing and assemble that drawn in Solidworks 2015 software. 77 SECTION 9: MODEL AND TECHNICAL DRAWING PICTURES OF OUR BUSINESS JET 9.1 FRONT VIEW 9.2 VIEW FROM LEFT 9.3 TOP VIEW 78 9.4 ISOMETRIC VIEW 79 REFERENCES 80