ACD506_Day 9& 10_Case Study 2
-
Upload
karthik-abhi -
Category
Documents
-
view
214 -
download
0
description
Transcript of ACD506_Day 9& 10_Case Study 2
-
M. S. Ramaiah University of Applied Sciences
1Faculty of Engineering & Technology
Session delivered by:
Dr. H. K. Narahari
Case Study 2 : Commercial Airliner
Session 9 & 10
-
M. S. Ramaiah University of Applied Sciences
2Faculty of Engineering & Technology
Case Study 2 : Commercial Airliner
Session Speaker
Dr. H.K. Narahari
-
M. S. Ramaiah University of Applied Sciences
3Faculty of Engineering & Technology
Requirement
Number of Passenger: 80
Range: 3000 nm = 5556 km
This is small transport jet plane category
-
M. S. Ramaiah University of Applied Sciences
4Faculty of Engineering & Technology
Preliminary Weight Estimation
-
M. S. Ramaiah University of Applied Sciences
5Faculty of Engineering & Technology
Empty Vs Takeoff Weight
-
M. S. Ramaiah University of Applied Sciences
6Faculty of Engineering & Technology
Typical Missions
-
M. S. Ramaiah University of Applied Sciences
7Faculty of Engineering & Technology
Mission Profile
-
M. S. Ramaiah University of Applied Sciences
8Faculty of Engineering & Technology
MP2
-
M. S. Ramaiah University of Applied Sciences
9Faculty of Engineering & Technology
MP3
-
M. S. Ramaiah University of Applied Sciences
10Faculty of Engineering & Technology
-
M. S. Ramaiah University of Applied Sciences
11Faculty of Engineering & Technology
MP4
-
M. S. Ramaiah University of Applied Sciences
12Faculty of Engineering & Technology
MP5
-
M. S. Ramaiah University of Applied Sciences
13Faculty of Engineering & Technology
Structure Weight Fraction
-
M. S. Ramaiah University of Applied Sciences
14Faculty of Engineering & Technology
Empty Weight Correlation (Airlines)
-
M. S. Ramaiah University of Applied Sciences
15Faculty of Engineering & Technology
Empty Weight Fraction (Airlines)
-
M. S. Ramaiah University of Applied Sciences
16Faculty of Engineering & Technology
Fuel Weight for MP1 1-2 Warm up and Takeoff >> (W1/W0) = 0.97
2-3 Climb >> (W2/W1) = 0.985
3-4 Cruise >>
therefore
Considered high bypass turbo jet engine cruising at 0.85 M with L/D = 0.866 (L/D) max
So, (W3/W2) = e-0.2154 = 0.8062
16
-
M. S. Ramaiah University of Applied Sciences
17Faculty of Engineering & Technology
Fuel Weight for MP1 4-5 Loiter and descent >>
According to FAA regulation an additional fuel for loitering at least for30 min has to be provided. In this case the additional time forendurance is taken as 45 min which includes both loiter and descent.Fuel ratio calculation for endurance is as follows
Therefore
So (W4/W3) = e-0.01874 = 0.9814
5-6 Landing phase >> (W5/W4) = 0.995
Therefore,
W5/W0 = (W1/WO) * (W2/W1) * (W3/W2) * (W4/W3) * (W5/W4) = 0.7521
-
M. S. Ramaiah University of Applied Sciences
18Faculty of Engineering & Technology
Weight Estimation
(Wf / Wo) = 1.06 * (1- W5/W0) = 0.2626
W payload = No of Passenger * (Wt of passenger + permissible baggage) = 80 * (80+40) = 9600 kg
W crew = 2 Pilot + 3 Cabin Crew = 5 * (80+40) = 600 kg
For W empty
We/Wo = A* WoC * Kvs
(We/Wo) = 1.0608 Wo-0.06
-
M. S. Ramaiah University of Applied Sciences
19Faculty of Engineering & Technology
Weight Estimation
So Total All up weight,
Wo = 22486.35/(0.7374-1.0608*Wo^-0.06)
After solving this by numerical method; Wo = 48957 kg
Approximately we can find from
graph
or table 600*80 = 48000 kg
OR
-
M. S. Ramaiah University of Applied Sciences
20Faculty of Engineering & Technology
Wing Design On basis of Performance spreadsheet
W/S required = 5000 N/m^2
Therefore for estimated W = 48957 kg
S = 96.053 m^2
Assume AR = 8.5
Therefore , b = 28.57 m
LE Sweep angle = 30 deg
Taper ratio = 0.2-0.3
AR = and
Cr = 5.17 m and Ct = 1.55 m
-
M. S. Ramaiah University of Applied Sciences
21Faculty of Engineering & Technology
Wing DesignAerofoil Selection:
Find Cruise Cl ;
L = Wavg = 0.5*(Wi+Wf) = 0.5**Cl*V^2*S
Therefore Cl = 0.465
But
The contribution of fuselage, tail and other components on overall lift has negative effect, So, Cl = Cl/0.95 = 0.489
3D wing error over 2D aerofoil = Cl = Cl/.9 = 0.54
NACA 6 series with10-12% t/C, required cruise Cl was not found so entire problem was worked out again with S = 120 m^2
With Cl = 0.41
-
M. S. Ramaiah University of Applied Sciences
22Faculty of Engineering & Technology
NACA 63-412 with Cl = 0.41 at 1.5deg
Clmax = 1.7
Cdmin = 0.0048
and stall is moderate
-
M. S. Ramaiah University of Applied Sciences
23Faculty of Engineering & Technology
Wing Design
NACA 63-412
3 D wing CAD model
Computational Domain
Final CAD drawings
-
M. S. Ramaiah University of Applied Sciences
24Faculty of Engineering & Technology
Powerplant Selection T/W = 0.35
For calculated AUW, Thrust required = 168 kN
Therefore two engines required with 85 kN thrust each
-
M. S. Ramaiah University of Applied Sciences
25Faculty of Engineering & Technology
Fuselage Layout Optimal aerodynamics, reducing aerodynamic drag Suppression of aerodynamic instability Comfortable and attractive seat design, placement, and
storage Space Safety features to deal with emergencies such as fires,
cabin depressurization, etc.; proper placement of emergency
exits, oxygen systems, etc. Easy handling for cargo loading and unloading, safe and
robust cargo hatches and doors Structural support for wing and tail forces acting in flight,
as well as for landing and ground operation forces
-
M. S. Ramaiah University of Applied Sciences
26Faculty of Engineering & Technology
Fuselage Layout Structurally optimized, saving weight while incorporating
protection against corrosion and fatigue Optimized flight deck, reducing pilot workload and protecting
against crew fatigue and intrusion by passengers Convenient size and placement of galleys, lavatories, and coat
racks Suppressed noise and vibration, providing a comfortable,
secure environment Control of cabin climate including air conditioning, heating,
and ventilation Providing housing for different sub-systems, including auxiliary
power units, hydraulic system, air conditioning, etc.
-
M. S. Ramaiah University of Applied Sciences
27Faculty of Engineering & Technology
Fuselage Layout : Major Dimensions
-
M. S. Ramaiah University of Applied Sciences
28Faculty of Engineering & Technology
Fuselage Layout : Why Circular A circle has the greatest cross-sectional area per unit
perimeter. The drag of a typical fuselage, which has a ratherlarge fineness ratio (l/d), is dominated by skin friction
A circle is strongest under internal pressure. At stratosphericcruising altitudes the outside pressure is 0.2 to 0.3 bar, whilethe internal pressure is maintained at that about 0.7 bar.Pressure difference across the thin skin of the cabin rangesfrom 0.4 to 0.5 atmospheres (40 to 50 kPa)
A circle more easily accommodates growth in Np in terms ofmanufacturing since cylindrical sections, called plugs, can bereasonably easily added to so-called stretched versions of agiven aircraft.
-
M. S. Ramaiah University of Applied Sciences
29Faculty of Engineering & Technology
Fuselage Layout Limited space outside the passenger compartment for
auxiliary systems and cargo. The passenger compartmentmust be located around a diameter of the circle for thegreatest width for seats and aisles.
Awkward circular sectors above and below the passengercompartment to house other items.
Modern designs have expanded the lower portion of thecircular cabin into a more rectangular cross-section in thevicinity of the wing root chord to accommodate more internalcarriage.
Cabin forward and aft of the wing root is maintained as acircular cross-section, and stretching will require plugs to beadded in these regions.
-
M. S. Ramaiah University of Applied Sciences
30Faculty of Engineering & Technology
Fuselage Typical Layout
-
M. S. Ramaiah University of Applied Sciences
31Faculty of Engineering & Technology
Fuselage Example Floor Plan
-
M. S. Ramaiah University of Applied Sciences
32Faculty of Engineering & Technology
Fuselage Drag Break down
-
M. S. Ramaiah University of Applied Sciences
33Faculty of Engineering & Technology
Fuselage Drag : Equation
-
M. S. Ramaiah University of Applied Sciences
34Faculty of Engineering & Technology
Fuselage Design
FAR rules have specified theminimum dimensions fordifferent class of passengerseats
The seat width considered forthe design is 500mm the seatpitch is 800mm and the aislewidth as 500mm.
-
M. S. Ramaiah University of Applied Sciences
35Faculty of Engineering & Technology
Fuselage Design
FAR rules state that duringemergency the plane needs to beevacuated within 90 second
Fuselage 3D CAD ModelFuselage Interior
Fuselage Seat Layout
Pilot Vision
-
M. S. Ramaiah University of Applied Sciences
36Faculty of Engineering & Technology
Empennage Design
The main function is to stabilize the aircraft in Pitch &Yaw andprovide control moments needed for maneuver and trim
In case of an engine failure the vertical tail must provideenough yaw moment to sustain the aircraft stable
About 70% of the aircrafts use a conventional tail
Symmetric Airfoil selected for the tail is NACA0012
-
M. S. Ramaiah University of Applied Sciences
37Faculty of Engineering & Technology
Tail Sizing
Pitching moment depends on wing chord and yawing momenton its span, Tail Volume Coefficients,
and
-
M. S. Ramaiah University of Applied Sciences
38Faculty of Engineering & Technology
Tail Sizing For twin engine general aviation
Ch = 0.80 and Cv = 0.07
Tail arm L is taken to 50% of the fuselage length = 18 m
Therefore Sh = 24.58 m^2 and Sv = 15.33 m^2
-
M. S. Ramaiah University of Applied Sciences
39Faculty of Engineering & Technology
Empennage Design The deflection of the control surface is at 25% of chord from trailing
edge and deflected to an angle of 35
NACA 0012 airfoil coordinates with deflected
control surface
39
-
M. S. Ramaiah University of Applied Sciences
40Faculty of Engineering & Technology
Assembly
-
M. S. Ramaiah University of Applied Sciences
41Faculty of Engineering & Technology
Conclusions A conceptual design of a commercial Jet liner to meet the
requirements to carry 80 passengers for range of 3000 nm ispresented
Selections of various aircraft systems and sub systems have beendone from available data, plots and thumb rules at conceptualdesign stage
There has been a focus on external aerodynamics (in this seminar)and practically nothing on structures
More design iterations have to be carried out after structuraldesign.
CFD analysis has to be carried out to find actual performance ofsystems and verify the required parameters
-
M. S. Ramaiah University of Applied Sciences
42Faculty of Engineering & Technology
At the end of this session the students would have understood basic requirements of :
Level Flight : Governing equations, Maximum and Minimum Velocities
Range and Endurance : Maximum range with and without a specified airspeed. Maximum endurance
Session Objectives
-
M. S. Ramaiah University of Applied Sciences
43Faculty of Engineering & Technology
Thank you !