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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
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