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Seminar-II
ROSHANSAH
USN :- 17AE60R01
M.Tech (1st Year)
Dept.of Aerospace Engg.
IndianInstituteof Technology
Kharagpur(IIT KGP)
PRELIMINARY DESIGN OF A 100 SEATER
AIRCRAFT
PHASES OF AIRPLANE DESIGN
Conceptual Design
-Variety of possible aircraft configuration arrangement, size, weight
& performance that meet the design specifications are sketched.
-New ideas & problem emerge as design is investigated.
Preliminary Design
-Initial stages of design
-Specialist in area of structure, landing gear ,control system will
design and analyze their portion of aircraft.
-Testing in area of aerodynamics, propulsion & stability and control
is initiated.
-One of key activity occur in it is lofting.
02/12/2018IIT KGP 2
Detail Design
-Begins with actual pieces to be fabricated are designed.
-Like; Precise design of each individual rib, spar and section of
skin takes place.
-Involves in Production design.
- specialist determine how the aircraft will be fabricated
starting with the smallest & simple sub-assemblies
and building up to the final assembly process.
- Testing effort intensifies and actual structure of aircraft is
fabricated and tested.
02/12/2018IIT KGP 3
Preliminary Design
β€’ Weight estimation
β€’ Wing design
β€’ Fuselage design
β€’ Drag polar
β€’ Maximum speed
β€’ Thrust requirement and selection of engine
β€’ Landing gear
β€’ Tail surfaces design
β€’ Airplane performance calculation
02/12/2018IIT KGP 4
Payload of 100 passenger
Range of 2000km
Cruise altitude = 12 km
Cruise Mach no = 0.7
Cruise speed = 743 km/hr =206.53 m/s
Landing distance = 1500 m
DESIGN REQUIREMENTS
02/12/2018IIT KGP 5
Crew members = 5
Weight of crew members (π‘Šπ‘)= 5*110 = 550 kg
Weight of passenger(π‘Šπ‘) = 100*110 = 11000 kg
Empty weight π‘Šπ‘’
Weight of fuel π‘Šπ‘“
Gross weight(π‘Š0) = π‘Šπ‘+π‘Šπ‘+π‘Šπ‘’+π‘Šπ‘“
= π‘Šπ‘ + π‘Šπ‘ +
π‘Š 𝑒
π‘Š0
π‘Š0 +
π‘Š 𝑓
π‘Š0
π‘Š0
Weight Estimation
02/12/2018IIT KGP 6
Weight Estimation
π‘Š0 =
π‘Šπ‘+π‘Šπ‘
1βˆ’
π‘Šπ‘’
π‘Š0
βˆ’
π‘Šπ‘“
π‘Š0
Wf = π‘Š0-W5
Fuel fraction estimate
π‘Š 𝑓
π‘Š0
= 1.06 1 βˆ’
π‘Š5
π‘Š0
π‘Š5
π‘Š0
=
π‘Š1
π‘Š0
βˆ—
π‘Š2
π‘Š1
βˆ—
π‘Š3
π‘Š2
βˆ—
π‘Š4
π‘Š3
βˆ—
π‘Š5
π‘Š4
02/12/2018IIT KGP 7
Historical mission segment weight fraction
Takeoff
π‘Š1
π‘Š0
) 0.97
Climb (
π‘Š2
π‘Š1
) 0.985
Landing (
π‘Š5
π‘Š4
) 0.995
Cruise
π‘Š3
π‘Š2
= 𝑒π‘₯𝑝
βˆ’π‘… 𝑐𝑑
π‘‰π‘π‘Ÿ
𝐿
𝐷 π‘π‘Ÿ
= 0.90147
π‘‰π‘π‘Ÿπ‘’π‘–π‘ π‘’ = 0.7 βˆ— 295.2 = 206.64 π‘š/𝑠
𝐿
𝐷 π‘šπ‘Žπ‘₯
= 15;
𝐿
𝐷 π‘π‘Ÿ
= 0.866
𝐿
𝐷 π‘šπ‘Žπ‘₯
= 13 ; 𝑐 𝑑 = 0.5/hr
02/12/2018IIT KGP 8
No. Aircraft Type 𝐿
𝐷
π‘šπ‘Žπ‘₯
1. Sailplane (glider) 20-35
2. Jet transport 12-20
3. Subsonicmilitary 8-11
4. Supersonicfighter 5-8
5. Helicopter 2-4
6. Homebuilt 6-14
7. Ultralight 8-15
02/12/2018IIT KGP 9
Loiter

π‘Š4
π‘Š3
= 𝑒π‘₯𝑝
βˆ’πΈ 𝑐𝑑
𝐿
𝐷 π‘šπ‘Žπ‘₯
= 0.9231
𝑐𝑑 = 0.4/hr and E =30 min.

π‘Š5
π‘Š0
=
π‘Š1
π‘Š0
βˆ—
π‘Š2
π‘Š1
βˆ—
π‘Š3
π‘Š2
βˆ—
π‘Š4
π‘Š3
βˆ—
π‘Š5
π‘Š4
= 0.791

π‘Š 𝑓
π‘Š0
= 0.22143
Weight Estimation
02/12/2018IIT KGP 10
π‘Šπ‘’
π‘Š0
= 1.02 π‘Š0
βˆ’0.06
π‘Š0 = 47350 kg
π‘Šπ‘’
π‘Š0
= 0.53466
Weight Estimation
02/12/2018IIT KGP 11
𝑆 𝐿 = 1500 m
ο‚§π‘‰π‘Ž =
𝑆 𝐿
0.3
0.5 = 128.0788 knot
where 𝑆 𝐿 is in feet and π‘‰π‘Ž is in knots
ο‚§π‘‰π‘Ž = 65.88 m/s
ο‚§π‘‰π‘ π‘‘π‘Žπ‘™π‘™ = π‘‰π‘Ž/1.3 = 50.68 m/s
 πΆπΏπ‘šπ‘Žπ‘₯= 2.5(using fowler flap)

π‘Š
𝑆 πΏπ‘Žπ‘›π‘‘
= 3933 N/m2

π‘Š
𝑆 𝑇𝑂
=
1
0.85
*
π‘Š
𝑆 πΏπ‘Žπ‘›π‘‘
= 4627 N/m2
οƒΌFinal choice of wing loading is =4346 N/m2 (+10 % variation in
π‘‰π‘ π‘‘π‘Žπ‘™π‘™
Choice Of Optimum Wing Loading
02/12/2018IIT KGP 12
Wing Design
Airfoil selection
-NASA SC(2) series airfoil to increase drag
divergence Mach no.
ο‚§πΆπΏπ‘π‘Ÿπ‘’π‘–π‘ π‘’ =
π‘Š
𝑆
π‘ž π‘π‘Ÿπ‘’π‘–π‘ π‘’
= 0.564
Airfoil thickness ratio
-At root - 15.2 % & At tip - 10.3%
-Average t/c = 14 %
Wing Sweep(quarter chord Sweep) = 25ΒΊ
Dihedral (Ξ“) = 5Β°
Wing twist = 3Β°
02/12/2018IIT KGP 13
Aspect Ratio(AR) = 9
 S =106.88 m2
Wing span = b = 𝐴𝑅. 𝑆
= 31.015 m
Taper Ratio(Ξ») = 0.24
Root chord(πΆπ‘Ÿ) =
2𝑆
𝑏 1+Ξ»
= 5.558 m
Tip chord(𝐢𝑑) =𝑐 π‘Ÿ. πœ† = 1.3339 m
Mean Aerodynamic Chord((𝒄)
𝒄 =
2
3
1+πœ†+πœ†2
1+πœ†
. 𝑐 π‘Ÿ = 3.877 m
Wing Design
02/12/2018IIT KGP 14
Fuselage comprises of nose and cockpit, passenger
compartment, and rear fuselage.
Fuselage Length(𝐿 𝑓) = π‘Žπ‘Š0
𝑏
; where a = 0.6 & b= 0.35
Fuselage Length(𝐿 𝑓) = 25.97m
Nose length(𝐿 π‘›π‘œπ‘ π‘’)
-𝐿 π‘›π‘œπ‘ π‘’ = 0.03𝐿 𝑓 = 0.7791m
Cockpit length = 2.5 m
Fuselage
02/12/2018IIT KGP 15
Fuselage
02/12/2018IIT KGP 16
Fuselage
Cabin parameters are chosen based on standards for similar airplanes
Parameter EconomyClass BusinessClass
Seat pitch(m) 0.8128 0.9652
Seat width(m) 0.508 0.5588
Aisle width(m) 0.5588 0.6096
Seats in a row 6 4
Numberofaisles 1 1
Max.height (m) 2.1 2.1
02/12/2018IIT KGP 17
Fuselage
Class No. of seats No. of rows Seat Pitch(m) Cabin
Length(m)
Economy 84 14 0.8128 11.3792
Business 16 4 0.9652 3.86
Cabin length = 11.3792+3.86 = 15.24 m
02/12/2018IIT KGP 18
Fuselage
Rear fuselage
Lrear= 0.25 Lf = 0.25 *25.97 = 6.501 m
Total fuselage length = 𝐿 π‘›π‘œπ‘ π‘’+ 𝐿 π‘π‘œπ‘π‘˜π‘π‘–π‘‘ + 𝐿 π‘π‘Žπ‘π‘–π‘› + 𝐿 π‘Ÿπ‘’π‘Žπ‘Ÿ
= 0.7791+2.5+ 15.24 + 6.501
= 25.02m
02/12/2018IIT KGP 19
Internal diameter of the cabin is calculated as:
- 𝐷𝑖 = Aisle width *1+(6*seat width) = 3.6068m =142inch
Structural thickness(in inches) is approximated by:
- t = 0.02*𝐷𝑖+1 = 3.84 inch = 0.097536m
The external diameter of the fuselage is obtained as:
- 𝐷0 =3.8018m= 149.68 inch
Fuselage
02/12/2018IIT KGP 20
Max Speed of aircraft
𝑀 π‘šπ‘Žπ‘₯ is taken as –
𝑀 π‘šπ‘Žπ‘₯ = π‘€π‘π‘Ÿπ‘’π‘–π‘ π‘’ + 0.04
𝑀 π‘šπ‘Žπ‘₯ = 0.74
π‘‰π‘š π‘Žπ‘₯ = 0.74*295.2 = 218.448 m/s
02/12/2018IIT KGP 21
Drag Polar
Assuming -
e(for wing)=0.85 ; Dihedral angle = 5
e(airplane) = 0.74 ; Aspect ratio(AR)=9
k=
1
Π𝑒𝐴𝑅
= 0.047794 ;
𝐿
𝐷 π‘šπ‘Žπ‘₯
=15
𝐢 𝐷0 = .0233
So the drag polar equation is 𝐢 𝐷 = 𝐢 𝐷0+kCL
2
𝐢 𝐷 = .0233+.047794CL
2
02/12/2018IIT KGP 22
Thrust Requirement
Weight at mid cruise

𝑇
π‘Š
= 0.5ρ𝑉2
π‘π‘Ÿπ‘’π‘–π‘ π‘’
𝑆
π‘Š
𝐢 𝐷0+
2πΎπ‘Š
𝑉2
π‘π‘Ÿπ‘’π‘–π‘ π‘’ 𝑆
ο‚§π‘Šπ‘šπ‘ = π‘Š2-0.5(π‘Š2-π‘Š3) = 45240.55- 0.5(45240.55-40783)
= 43011.77 kg

π‘Š π‘šπ‘
𝑆
= 3947.84. N/m2
Total thrust produced (Tcruise ) =62.96891KN
Tcruise each engine = 31.484 KN
02/12/2018IIT KGP 23
Thrust Requirement

𝑇 𝑆𝐿
π‘Š
Οƒ = 0.5ρ𝑉2
π‘šπ‘Žπ‘₯
𝑆
π‘Š
𝐢 𝐷0+
2πΎπ‘Š
ρ 𝑉2
max 𝑆
𝑇𝑆𝐿= 109.80KN
Thrust require maximum is 109.80KN
Based on the above calculations
ROLLS-ROYCE BR715/710 Turbofan
is selected.
Specification:-
β€’ Thrust range = 14000lb to 21500 lb.
β€’ 20% improved fuel burn
β€’ Fan diameter 58 inch and a two-stage
booster driven by a three-stage LP turbine.
β€’ single low emissions annular combustor with 20 burners.
02/12/2018IIT KGP 24
Landing Gear
 A conventional Tricycle landing gear is chosen based on the
trend followed by similar airplanes.
Parameter Value
Wheel base (in m) 13.2
Track length (in m) 5.8
Turning radius(in m) 19.3
02/12/2018IIT KGP 25
Tail Surfaces
Parameters HorizontalTail Vertical Tail
Area Ratio 0.31 0.21
AspectRatio 5 1.7
TaperRatio 0.26 0.31
Sh=0.31*106.88=33.1338 m2 bh = 12.87m
Sv=0.21*106.88=22.44 m2
bv = 6.1764 m
02/12/2018IIT KGP 26
Tail Arm
The values of the tail arm are chosen based on the data
collection as:
Lh = 45 % of Lf &
 Lv = 42% of Lf
Lh = 0.45 x 25.02=11.259 m
Lv = 0.42 x 25.02= 10.5084m
Vh =
π‘†β„Ž πΏβ„Ž
𝑆𝐢
= 0.9
Vv =
𝑆 𝑉 𝐿 𝑉
𝑆𝐢
= .0.569
02/12/2018IIT KGP 27
Three view drawing of Boeing 737-200
Three view Drawing
02/12/2018IIT KGP 28
[1] D.P. Raymer β€œAircraft design: a conceptual approach”AIAAeducational Series, Fourth
edition (2006).
[2] Harris C.H. β€œNASA supercritical airfoils – A matrix of family- related airfoils” NASA
TP 2969 , March 1990
[3] Jenkinson L. R., Simpkin P. and Rhodes D. β€œCivil jet aircraft design”Arnold,(1999).
[4]Introduction to Flight by John D.Anderson,2nd edition.
[5]Theory of wing sections by Ira. H. Abbott, Dover edition.
REFERENCES :-
02/12/2018IIT KGP 29

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Preliminary Design of 100 Seater Aircraft

  • 1. Seminar-II ROSHANSAH USN :- 17AE60R01 M.Tech (1st Year) Dept.of Aerospace Engg. IndianInstituteof Technology Kharagpur(IIT KGP) PRELIMINARY DESIGN OF A 100 SEATER AIRCRAFT
  • 2. PHASES OF AIRPLANE DESIGN Conceptual Design -Variety of possible aircraft configuration arrangement, size, weight & performance that meet the design specifications are sketched. -New ideas & problem emerge as design is investigated. Preliminary Design -Initial stages of design -Specialist in area of structure, landing gear ,control system will design and analyze their portion of aircraft. -Testing in area of aerodynamics, propulsion & stability and control is initiated. -One of key activity occur in it is lofting. 02/12/2018IIT KGP 2
  • 3. Detail Design -Begins with actual pieces to be fabricated are designed. -Like; Precise design of each individual rib, spar and section of skin takes place. -Involves in Production design. - specialist determine how the aircraft will be fabricated starting with the smallest & simple sub-assemblies and building up to the final assembly process. - Testing effort intensifies and actual structure of aircraft is fabricated and tested. 02/12/2018IIT KGP 3
  • 4. Preliminary Design β€’ Weight estimation β€’ Wing design β€’ Fuselage design β€’ Drag polar β€’ Maximum speed β€’ Thrust requirement and selection of engine β€’ Landing gear β€’ Tail surfaces design β€’ Airplane performance calculation 02/12/2018IIT KGP 4
  • 5. Payload of 100 passenger Range of 2000km Cruise altitude = 12 km Cruise Mach no = 0.7 Cruise speed = 743 km/hr =206.53 m/s Landing distance = 1500 m DESIGN REQUIREMENTS 02/12/2018IIT KGP 5
  • 6. Crew members = 5 Weight of crew members (π‘Šπ‘)= 5*110 = 550 kg Weight of passenger(π‘Šπ‘) = 100*110 = 11000 kg Empty weight π‘Šπ‘’ Weight of fuel π‘Šπ‘“ Gross weight(π‘Š0) = π‘Šπ‘+π‘Šπ‘+π‘Šπ‘’+π‘Šπ‘“ = π‘Šπ‘ + π‘Šπ‘ + π‘Š 𝑒 π‘Š0 π‘Š0 + π‘Š 𝑓 π‘Š0 π‘Š0 Weight Estimation 02/12/2018IIT KGP 6
  • 7. Weight Estimation π‘Š0 = π‘Šπ‘+π‘Šπ‘ 1βˆ’ π‘Šπ‘’ π‘Š0 βˆ’ π‘Šπ‘“ π‘Š0 Wf = π‘Š0-W5 Fuel fraction estimate π‘Š 𝑓 π‘Š0 = 1.06 1 βˆ’ π‘Š5 π‘Š0 π‘Š5 π‘Š0 = π‘Š1 π‘Š0 βˆ— π‘Š2 π‘Š1 βˆ— π‘Š3 π‘Š2 βˆ— π‘Š4 π‘Š3 βˆ— π‘Š5 π‘Š4 02/12/2018IIT KGP 7
  • 8. Historical mission segment weight fraction Takeoff π‘Š1 π‘Š0 ) 0.97 Climb ( π‘Š2 π‘Š1 ) 0.985 Landing ( π‘Š5 π‘Š4 ) 0.995 Cruise π‘Š3 π‘Š2 = 𝑒π‘₯𝑝 βˆ’π‘… 𝑐𝑑 π‘‰π‘π‘Ÿ 𝐿 𝐷 π‘π‘Ÿ = 0.90147 π‘‰π‘π‘Ÿπ‘’π‘–π‘ π‘’ = 0.7 βˆ— 295.2 = 206.64 π‘š/𝑠 𝐿 𝐷 π‘šπ‘Žπ‘₯ = 15; 𝐿 𝐷 π‘π‘Ÿ = 0.866 𝐿 𝐷 π‘šπ‘Žπ‘₯ = 13 ; 𝑐 𝑑 = 0.5/hr 02/12/2018IIT KGP 8
  • 9. No. Aircraft Type 𝐿 𝐷 π‘šπ‘Žπ‘₯ 1. Sailplane (glider) 20-35 2. Jet transport 12-20 3. Subsonicmilitary 8-11 4. Supersonicfighter 5-8 5. Helicopter 2-4 6. Homebuilt 6-14 7. Ultralight 8-15 02/12/2018IIT KGP 9
  • 10. Loiter  π‘Š4 π‘Š3 = 𝑒π‘₯𝑝 βˆ’πΈ 𝑐𝑑 𝐿 𝐷 π‘šπ‘Žπ‘₯ = 0.9231 𝑐𝑑 = 0.4/hr and E =30 min.  π‘Š5 π‘Š0 = π‘Š1 π‘Š0 βˆ— π‘Š2 π‘Š1 βˆ— π‘Š3 π‘Š2 βˆ— π‘Š4 π‘Š3 βˆ— π‘Š5 π‘Š4 = 0.791  π‘Š 𝑓 π‘Š0 = 0.22143 Weight Estimation 02/12/2018IIT KGP 10
  • 11. π‘Šπ‘’ π‘Š0 = 1.02 π‘Š0 βˆ’0.06 π‘Š0 = 47350 kg π‘Šπ‘’ π‘Š0 = 0.53466 Weight Estimation 02/12/2018IIT KGP 11
  • 12. 𝑆 𝐿 = 1500 m ο‚§π‘‰π‘Ž = 𝑆 𝐿 0.3 0.5 = 128.0788 knot where 𝑆 𝐿 is in feet and π‘‰π‘Ž is in knots ο‚§π‘‰π‘Ž = 65.88 m/s ο‚§π‘‰π‘ π‘‘π‘Žπ‘™π‘™ = π‘‰π‘Ž/1.3 = 50.68 m/s  πΆπΏπ‘šπ‘Žπ‘₯= 2.5(using fowler flap)  π‘Š 𝑆 πΏπ‘Žπ‘›π‘‘ = 3933 N/m2  π‘Š 𝑆 𝑇𝑂 = 1 0.85 * π‘Š 𝑆 πΏπ‘Žπ‘›π‘‘ = 4627 N/m2 οƒΌFinal choice of wing loading is =4346 N/m2 (+10 % variation in π‘‰π‘ π‘‘π‘Žπ‘™π‘™ Choice Of Optimum Wing Loading 02/12/2018IIT KGP 12
  • 13. Wing Design Airfoil selection -NASA SC(2) series airfoil to increase drag divergence Mach no. ο‚§πΆπΏπ‘π‘Ÿπ‘’π‘–π‘ π‘’ = π‘Š 𝑆 π‘ž π‘π‘Ÿπ‘’π‘–π‘ π‘’ = 0.564 Airfoil thickness ratio -At root - 15.2 % & At tip - 10.3% -Average t/c = 14 % Wing Sweep(quarter chord Sweep) = 25ΒΊ Dihedral (Ξ“) = 5Β° Wing twist = 3Β° 02/12/2018IIT KGP 13
  • 14. Aspect Ratio(AR) = 9  S =106.88 m2 Wing span = b = 𝐴𝑅. 𝑆 = 31.015 m Taper Ratio(Ξ») = 0.24 Root chord(πΆπ‘Ÿ) = 2𝑆 𝑏 1+Ξ» = 5.558 m Tip chord(𝐢𝑑) =𝑐 π‘Ÿ. πœ† = 1.3339 m Mean Aerodynamic Chord((𝒄) 𝒄 = 2 3 1+πœ†+πœ†2 1+πœ† . 𝑐 π‘Ÿ = 3.877 m Wing Design 02/12/2018IIT KGP 14
  • 15. Fuselage comprises of nose and cockpit, passenger compartment, and rear fuselage. Fuselage Length(𝐿 𝑓) = π‘Žπ‘Š0 𝑏 ; where a = 0.6 & b= 0.35 Fuselage Length(𝐿 𝑓) = 25.97m Nose length(𝐿 π‘›π‘œπ‘ π‘’) -𝐿 π‘›π‘œπ‘ π‘’ = 0.03𝐿 𝑓 = 0.7791m Cockpit length = 2.5 m Fuselage 02/12/2018IIT KGP 15
  • 17. Fuselage Cabin parameters are chosen based on standards for similar airplanes Parameter EconomyClass BusinessClass Seat pitch(m) 0.8128 0.9652 Seat width(m) 0.508 0.5588 Aisle width(m) 0.5588 0.6096 Seats in a row 6 4 Numberofaisles 1 1 Max.height (m) 2.1 2.1 02/12/2018IIT KGP 17
  • 18. Fuselage Class No. of seats No. of rows Seat Pitch(m) Cabin Length(m) Economy 84 14 0.8128 11.3792 Business 16 4 0.9652 3.86 Cabin length = 11.3792+3.86 = 15.24 m 02/12/2018IIT KGP 18
  • 19. Fuselage Rear fuselage Lrear= 0.25 Lf = 0.25 *25.97 = 6.501 m Total fuselage length = 𝐿 π‘›π‘œπ‘ π‘’+ 𝐿 π‘π‘œπ‘π‘˜π‘π‘–π‘‘ + 𝐿 π‘π‘Žπ‘π‘–π‘› + 𝐿 π‘Ÿπ‘’π‘Žπ‘Ÿ = 0.7791+2.5+ 15.24 + 6.501 = 25.02m 02/12/2018IIT KGP 19
  • 20. Internal diameter of the cabin is calculated as: - 𝐷𝑖 = Aisle width *1+(6*seat width) = 3.6068m =142inch Structural thickness(in inches) is approximated by: - t = 0.02*𝐷𝑖+1 = 3.84 inch = 0.097536m The external diameter of the fuselage is obtained as: - 𝐷0 =3.8018m= 149.68 inch Fuselage 02/12/2018IIT KGP 20
  • 21. Max Speed of aircraft 𝑀 π‘šπ‘Žπ‘₯ is taken as – 𝑀 π‘šπ‘Žπ‘₯ = π‘€π‘π‘Ÿπ‘’π‘–π‘ π‘’ + 0.04 𝑀 π‘šπ‘Žπ‘₯ = 0.74 π‘‰π‘š π‘Žπ‘₯ = 0.74*295.2 = 218.448 m/s 02/12/2018IIT KGP 21
  • 22. Drag Polar Assuming - e(for wing)=0.85 ; Dihedral angle = 5 e(airplane) = 0.74 ; Aspect ratio(AR)=9 k= 1 Π𝑒𝐴𝑅 = 0.047794 ; 𝐿 𝐷 π‘šπ‘Žπ‘₯ =15 𝐢 𝐷0 = .0233 So the drag polar equation is 𝐢 𝐷 = 𝐢 𝐷0+kCL 2 𝐢 𝐷 = .0233+.047794CL 2 02/12/2018IIT KGP 22
  • 23. Thrust Requirement Weight at mid cruise  𝑇 π‘Š = 0.5ρ𝑉2 π‘π‘Ÿπ‘’π‘–π‘ π‘’ 𝑆 π‘Š 𝐢 𝐷0+ 2πΎπ‘Š 𝑉2 π‘π‘Ÿπ‘’π‘–π‘ π‘’ 𝑆 ο‚§π‘Šπ‘šπ‘ = π‘Š2-0.5(π‘Š2-π‘Š3) = 45240.55- 0.5(45240.55-40783) = 43011.77 kg  π‘Š π‘šπ‘ 𝑆 = 3947.84. N/m2 Total thrust produced (Tcruise ) =62.96891KN Tcruise each engine = 31.484 KN 02/12/2018IIT KGP 23
  • 24. Thrust Requirement  𝑇 𝑆𝐿 π‘Š Οƒ = 0.5ρ𝑉2 π‘šπ‘Žπ‘₯ 𝑆 π‘Š 𝐢 𝐷0+ 2πΎπ‘Š ρ 𝑉2 max 𝑆 𝑇𝑆𝐿= 109.80KN Thrust require maximum is 109.80KN Based on the above calculations ROLLS-ROYCE BR715/710 Turbofan is selected. Specification:- β€’ Thrust range = 14000lb to 21500 lb. β€’ 20% improved fuel burn β€’ Fan diameter 58 inch and a two-stage booster driven by a three-stage LP turbine. β€’ single low emissions annular combustor with 20 burners. 02/12/2018IIT KGP 24
  • 25. Landing Gear  A conventional Tricycle landing gear is chosen based on the trend followed by similar airplanes. Parameter Value Wheel base (in m) 13.2 Track length (in m) 5.8 Turning radius(in m) 19.3 02/12/2018IIT KGP 25
  • 26. Tail Surfaces Parameters HorizontalTail Vertical Tail Area Ratio 0.31 0.21 AspectRatio 5 1.7 TaperRatio 0.26 0.31 Sh=0.31*106.88=33.1338 m2 bh = 12.87m Sv=0.21*106.88=22.44 m2 bv = 6.1764 m 02/12/2018IIT KGP 26
  • 27. Tail Arm The values of the tail arm are chosen based on the data collection as: Lh = 45 % of Lf &  Lv = 42% of Lf Lh = 0.45 x 25.02=11.259 m Lv = 0.42 x 25.02= 10.5084m Vh = π‘†β„Ž πΏβ„Ž 𝑆𝐢 = 0.9 Vv = 𝑆 𝑉 𝐿 𝑉 𝑆𝐢 = .0.569 02/12/2018IIT KGP 27
  • 28. Three view drawing of Boeing 737-200 Three view Drawing 02/12/2018IIT KGP 28
  • 29. [1] D.P. Raymer β€œAircraft design: a conceptual approach”AIAAeducational Series, Fourth edition (2006). [2] Harris C.H. β€œNASA supercritical airfoils – A matrix of family- related airfoils” NASA TP 2969 , March 1990 [3] Jenkinson L. R., Simpkin P. and Rhodes D. β€œCivil jet aircraft design”Arnold,(1999). [4]Introduction to Flight by John D.Anderson,2nd edition. [5]Theory of wing sections by Ira. H. Abbott, Dover edition. REFERENCES :- 02/12/2018IIT KGP 29