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



Case Study 2 : Commercial Airliner

Session Speaker

Dr. H.K. Narahari



Requirement

Number of Passenger: 80

Range: 3000 nm = 5556 km

This is small transport jet plane category



Preliminary Weight Estimation



Empty Vs Takeoff Weight



Typical Missions



Mission Profile



MP2



MP3



MP4



MP5



Structure Weight Fraction



Empty Weight Correlation (Airlines)



Empty Weight Fraction (Airlines)



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



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



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



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



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



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



NACA 63-412 with Cl = 0.41 at 1.5deg

Clmax = 1.7

Cdmin = 0.0048

and stall is moderate



Wing Design

NACA 63-412

3 D wing CAD model

Computational Domain

Final CAD drawings



Powerplant Selection T/W = 0.35

For calculated AUW, Thrust required = 168 kN

Therefore two engines required with 85 kN thrust each



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



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.



Fuselage Layout : Major Dimensions



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.



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.



Fuselage Typical Layout



Fuselage Example Floor Plan



Fuselage Drag Break down



Fuselage Drag : Equation



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.



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



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



Tail Sizing

Pitching moment depends on wing chord and yawing momenton its span, Tail Volume Coefficients,

and



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



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



Assembly



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



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



Thank you !

ACD506_Day 9& 10_Case Study 2

Documents

Transcript of ACD506_Day 9& 10_Case Study 2