1

Analysis of Airbus A320neo

John Connolly, Robert Lew, Michael Marcolini, Scot Surprenant

Mechanical Engineering (BS) Candidates, Wentworth Institute of Technology, Boston, MA

Technical Advisor: Haifa El-Sadi, Ph. D

Associate Professor, Wentworth Institute of Technology, Boston, MA

This report is an elementary analysis of the Airbus A320neo. The A320neo was chosen as

the subject because it is a widely used midsize commercial aircraft. Using basic published

values of the aircraft’s specifications and general equations for aircraft design, students

completed calculations estimating the weight and various subsystem geometries. Geometric

values were compared with the actual measurements to ensure accuracy of the estimates.

Aerodynamic coefficients were assessed for all surfaces of the aircraft. A simplified model

was generated and a CFD analysis was completed using the using SolidWorks Flow

Simulation. The simulation results were similar to the results of the calculations.

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I. Abstract

Students were tasked with completing an elementary analysis on the basic parameters and

physical characteristics on a modern aircraft. This project served as an introduction to the

fundamentals of airplane design learned in an aerodynamics course. The A320neo was chosen as

the subject because it is a widely used midsize commercial aircraft. Using basic published values

of the aircraft’s specifications and general equations for aircraft design, students completed

calculations estimating the weight and various subsystem geometries. Geometric values were

compared with the actual measurements to ensure accuracy of the estimates. Aerodynamic

coefficients were assessed for all surfaces of the aircraft. A simplified model was generated and a

CFD analysis was completed using the using SolidWorks Flow Simulation. Values calculated for

the geometries of the wing, fuselage and stabilizers fell within a 5% error of the advertised

values. The coefficient of lift and drag calculations were similar to the simulation results. These

values are not published so they could not be verified from an outside source.

II. Introduction

The Airbus A320Neo is a jet airliner in the A320 family, produced by Airbus. The A320neo

(new engine option) is one of many upgrades introduced by Airbus to help maintain its A320

product line’s position as the world’s most advanced and fuel-efficient single-aisle aircraft family.

Capable of seating up to 180 passengers and traveling 6300 km the A320neo is one of the most

efficient jets in its class. The first A320neo entered commercial service in January 2016 with a

PW1100G-JM engine and with a LEAP1A engine option in July 2016.

The purpose of this report is due analyze the performance and efficiency of the Airbus A320neo.

Dimensions and performance will be validated using aerodynamic calculations and computational

fluid dynamics (CFD) in SolidWorks. The major components of the A320neo are shown below.

The A320neo is most efficient at its ceiling of 39,000 ft, where it flies at 0.79 Mach. The

diagram below depicts the efficacy of the A320 through its stages of flight. This report will analyze

the plane’s performance at this most efficient flight stage.

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III. Weight of Aircraft

Data must be gathered about the Airbus 320neo to determine the gross weight (𝑊0) of the

aircraft. The components of the aircraft weight are broken down as follows:

Variable Component

𝑊𝑐𝑟𝑒𝑤 Crew weight

𝑊𝑝𝑎𝑥 Passenger weight

𝑊𝑝−𝑙𝑜𝑎𝑑 Payload weight

𝑊𝑓 Fuel weight

𝑊𝑒 Empty aircraft weight

𝑾𝟎 Total: Gross takeoff weight

A regular crew consists of four people with a typical passenger limit of 180 people. Passengers

can be assumed to be 180 lbs. Each passenger and crewmember are allotted 50 lbs of additional

payload (luggage, carry-on, refreshments).

Count (#) Average

Weight (lbs)

Subtotal

Weight (lbs) Variable

Crewmembers 4 180 720 𝑾𝒄𝒓𝒆𝒘

Passengers 190 180 34200 𝑾𝒑𝒂𝒙

Payload 194 50 9700 𝑾𝒑−𝒍𝒐𝒂𝒅

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The empty weight ratio must be calculated using the following equation:

𝑊𝑒

𝑊0= 𝐴 ∗ 𝑊0

𝑐 ∗ 𝑘𝑣𝑠

Where A and c are constants found in historical texts and 𝑘𝑢𝑠 is the variable sweep constant.

The values are seen in the chart below. Based on an initial guess of 𝑊0 = 129380 lb, the initial

empty weight ratio 𝑾𝒆

𝑾𝟎= 𝟎. 𝟒𝟖𝟐𝟒.

Variable Value

𝐴 1.02

𝑐 -0.0636

𝑘𝑣𝑠 1

To calculate the fuel weight fraction (𝑊𝑓

𝑊0), a weight fraction based on the flight stages, must be

calculated.

𝑊𝑓

𝑊0= 1.06 ∗ (1 −

𝑊𝑥

𝑊0) , where

𝑊𝑥

𝑊0=

𝑊1

𝑊0∗

𝑊2

𝑊1∗

𝑊3

𝑊2∗

𝑊4

𝑊3∗

𝑊5

𝑊4.

The stages of flight are given as follows:

Stage

1 Warmup

2 Climb

3 Cruise

4 Loiter

5 Landing

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After each stage, a weight fraction is calculated for the equation above. Some values are given,

others are calculated based on the aircraft’s attributes.

Stage Weight

Fraction Expression Equation Value

𝑊1

𝑊0 Given: 0.970 0.970

𝑊2

𝑊1 Given: 0.985 0.985

𝑊3

𝑊2

= 𝑒[−

𝑅∗𝐶

𝑉∗(𝐿𝐷⁄ )

]

0.790

R (range) 2.28E+07 ft

C (Specific Fuel Consumption) 0.000139 1/s

V (Cruise velocity) 765 ft/s

L/D (Lift/Drag ratio) 17.5

𝑊4

𝑊3

= 𝑒[−

𝐸∗𝐶

(𝐿𝐷⁄ )

]

0.992411

E (Endurance time) 1200 s

C (Specific Fuel Consumption) 0.000111 1/s

𝑊5

𝑊4 Given: 0.995 0.995

𝑊𝑥

𝑊0 =

𝑊1

𝑊0∗

𝑊2

𝑊1∗

𝑊3

𝑊2∗

𝑊4

𝑊3∗

𝑊5

𝑊4 0.744

𝑾𝒇

𝑾𝟎 = 𝟏. 𝟎𝟔 ∗ (𝟏 −

𝑾𝒙

𝑾𝟎) 0.270

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The values from each of the calculations above are substituted in the equation below.

𝑊0 =𝑊𝑐𝑟𝑒𝑤 + 𝑊𝑝−𝑙𝑜𝑎𝑑 + 𝑊𝑝𝑎𝑥

1 − (𝑊𝑓 − 𝑊0) − (𝑊𝑒

𝑊0)

Once a new gross weight is calculated, the empty weight fraction is recalculated with the new

value. This process can be iterated to find a convergent solution for 𝑊0.

Iteration 𝑊0𝑖

𝑊𝑒

𝑊0 𝑊0

𝑓

0 129380 0.482487 180573

1 180573 0.472364 173467

2 173467 0.473572 174285

3 174285 0.473430 174189

4 174188.9 0.473447 174200.2

5 174200.2 0.473445 174198.9

6 174198.9 0.473445 174199.1

7 174199.1 0.473445 174199.1

8 174199.1 0.473445 174199.1

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The percent error between the posted value and the calculated value, for the gross takeoff

weight, the empty weight ratio, and the fuel weight ratio are listed below.

Posted Calculated Error

Gross Takeoff

Weight (𝑊0) 174000 174199.1 0.114 %

Empty Weight

Fraction (𝑊𝑒

𝑊0)

0.473191 0.473445 0.054 %

Fuel Weight

Fraction (𝑊𝑓

𝑊0)

0.270372 0.270411 0.014 %

The calculations performed in this section are extremely close to that of the actual A320neo

(only 0.1% error from actual gross takeoff weight). These calculations give insight to the

required weight of the aircraft based on several performance attributes: payload, cruising speed,

range, among others. Altering these attributes would change the required weights for the fuel and

the aircraft itself.

IV. Fuselage and Wing Dimensions

Using the equation and given values from table below, fuselage length can be calculated.

Length = 𝑎 ∗ 𝑊0𝐶

Aircraft Type a c

Jet Transport 0.67 0.43

Calculated Length = 0.67 ∗ 1740000.43 = 𝟏𝟐𝟎. 𝟐 ft

Calculated

Length

Actual

Length

Percent

Error

120.2 ft 123.3 ft 2.51%

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Using the equations, 𝐿𝑓

𝑑𝑓= 9.5, 𝑆𝐹 =

𝜋∗𝑑𝑓2

4,

Variable Value

Length (Lf) 123.3 ft

Diameter (df) 12.96 ft

Fuselage Area (SF) 131.9 ft2

The data in the following chart was found in an Aircraft Characteristics and Maintenance

document for the Airbus A320 (drawing shown above).

Variable Value Dimension

Span (b) 117.45 ft

Wing Area (s) 1460 ft2

Aspect Ratio (AR) 9.45 N/A

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The calculations in the chart below map out the geometry of the wing. All calculated values

are within five percent error of the actual dimensions of the plane.

Wing Calculations

Variable Eq Value Dimension

Aspect Ratio (AR) 𝑏2 ∗ 𝑠 9.45

Root chord (𝐶𝑟𝑜𝑜𝑡) 2 ∗ 𝑠

𝑏(1 + λ) 19.57 ft

Tip chord (𝐶𝑡𝑖𝑝) λ ∗ 𝐶𝑟𝑜𝑜𝑡 5.29 ft

Mean Aero Chord (𝑐) 2

3∗ 𝐶𝑟𝑜𝑜𝑡 ∗

(1 + λ +λ2)

(1 + λ) 13.80 ft

��

23.74 ft

Sweep Angle (𝛬𝐿𝐸) tan−1 (𝑐𝑟𝑜𝑜𝑡

𝑏2⁄

) 18.43 deg

Quarter Chord Line

Angle (𝛬𝑐

4) tan−1 (

𝑐4⁄

��) 8.27 deg

The following properties are used to calculate the lift, drag, and moment coefficient for the plane.

Airfoil Calcs

Calcs Value Dimension

𝐴𝑖𝑟 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 0.000738 slugs/ft3

Velocity 765 ft/s

q 215.95 slugs/(ft*s2)

Dynamic Viscosity 2.995E-07

Reynolds Number 26x10^6

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The airfoil used in this analysis is the NACA 63-412. It resembles the A320neo’s supercritical

airfoil in terms of thickness and camber. The charts below indicate the performance of the airfoil

at the calculated Reynolds number (~25,000,000).

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The charts above are used to find lift and drag coefficients at various angles of attack. Then, the

equations used for lift, drag, and moment coefficient are applied:

𝐶𝐿 =𝐿

𝑞 ∗ 𝑠

𝐶𝐷 =𝐷

𝑞 ∗ 𝑠

𝐶𝑚 =𝑚

𝑞 ∗ 𝑠 ∗ 𝑐𝑏𝑎𝑟

Angle of

Attack

Lift

Coefficient

(cL)

Drag

Coefficient

(cD)

Moment

Coefficient

(cm,ac)

Lift (lbf)

Drag (lbf)

0 0.3401 0.0060 -0.0785 107228 1882

8 1.1640 0.0146 -0.0703 366991 4597

12 1.4237 0.0243 -0.0464 448870 7649

18 1.4642 0.0910 -0.0462 461639 28694

Dimensions from the A320neo’s maintenance manual were used to calculate more detailed

variables and dimensions to describe the fuselage and wing’s characteristics. A six-digit NACA

airfoil was applied as a substitute for the A320neo’s supercritical airfoil. This allowed access to a

library of lift and drag coefficients for high Reynolds number applications. Using these

coefficients, total lift and drag due to the wings could be calculated.

The total lift and drag makes sense with reality. The calculations show an angle of attack

between 0-8˚ would suffice to maintain level flight (actual angle of incidence is approximately

4˚). An angle of attack of 12˚ provides a very high lift to drag ratio. At 18˚, the airfoil begins to

separate, and although lift increases marginally, drag increases dramatically.

Flap Dimensions:

The flap calculations are shown in the chart below. The dimensions of the flaps are derived

from the wing dimensions found above. These calculated values fall very close to the published

values of the actual dimensions. The stall speed and takeoff speed are also within a 5% margin of

error.

Name Formula Value Dimension

Flap Chord 𝐶𝑓 = 0.36𝐶 7.04 ft

Flap Span 𝑏𝑓 = 0.67𝐵 78.6 ft

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Stall Speed 𝑉𝑠 = √ 2𝑊

32.2(𝑠 ∗ 𝜌 ∗ 𝐶𝐿) 218.6 ft/s

Take off Velocity 𝑉𝑇𝑂 = 1.2𝑉𝑠 264.4 ft/s

Takeoff Wight 𝑊𝑇𝑂 = 0.97𝑊0 16627 lb.

Coefficient of Lift 𝐶𝐿𝑇𝑂 =2𝑊𝑇𝑂

2𝜌𝑉𝑇𝑂𝑠 1.41

Washout:

The A320neo features a washout angle of 4.11˚. This gives the aircraft stability because at the

tip, the lower angle of incidence decreases the roll moment at the tip. In other words, the

significant lift area is at the center of the aircraft.

Solving for Critical Mach Number (Graphical Solution):

𝑣 = 𝑀 ∗ √𝑘 ∗ 𝑅 ∗ 𝑇𝑐𝑟𝑢𝑖𝑠𝑒

𝐶𝑃0 = 1 − (𝑣

𝑣𝑐𝑟𝑢𝑖𝑠𝑒)

2

𝐶𝑃 = (𝐶𝑃0

√1 − 𝑀∞2

)

2

−

+

−+=

−

11

)1(22)1/(

2

2,

kk

crpk

Mk

kMC

13

𝑴𝒄𝒓 = 𝟎. 𝟖𝟓𝟏

Solving for Critical Mach Number (Analytical Solution):

Mcr 2

0,

1 −M

Cp

−

+

−+−

11

)1(22)1/(

2

2

kk

k

Mk

kM

Percent

Difference

0.847 -0.27930 -0.30938 2.56%

0.848 -0.28526 -0.30691 1.83%

0.849 -0.29126 -0.30445 1.11%

0.850 -0.29732 -0.30199 0.39%

0.851 -0.30342 -0.29954 0.32%

0.852 -0.30957 -0.29710 1.03%

0.853 -0.31576 -0.29467 1.73%

0.854 -0.32201 -0.29224 2.42%

0.855 -0.32831 -0.28982 3.11%

The analytical solution approach was like the graphical solution in that the pressure

coefficient (including the compressibility correction) is compared to the critical pressure

coefficient value. The point at which the values approach each other is the point at which the

critical Mach number is approached. Both methods found the critical Mach number to be 0.851

because they are calculated using the same equations.

This critical Mach number makes sense because our plane cruises at 0.78 Mach. At anything

higher than 0.78, shockwaves would likely occur and drastically increases drag. 0.78M is an

efficient flight mode for the A320neo.

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V. Stabilizers

Vertical Stabilizer

To create initial dimensions, the attributes of the Boeing 737-100 will be used as reference.

This is an appropriate comparison because the 737’s takeoff-weight matches that of the

A320neo: 174,200 lb. (0% difference).

Tail Characteristic

Boeing

737-100

Value

𝑆𝑉 𝑆𝑤𝑖𝑛𝑔⁄ 0.27

𝐴𝑅𝑉 1.88

𝛬𝑉 35°

𝑆𝑉 𝑆𝑤𝑖𝑛𝑔⁄ = 0.27

𝑆𝑉 = 0.27 ∗ 𝑆𝑤𝑖𝑛𝑔

𝑆𝑉 = 0.27 ∗ 1460

𝑺𝑽 = 𝟑𝟗𝟒. 𝟐 ft2

𝐴𝑅𝑉 =𝑏𝑉

2

𝑆𝑉

𝑏𝑉 = √𝐴𝑅𝑉 ∗ 𝑆𝑉

𝑏𝑉 = √1.88 ∗ 394.2

𝒃𝑽 = 𝟐𝟕. 𝟐𝟐 ft

𝑆𝑉 = 𝑏𝑉 ∗ 𝑐��

𝑐�� =𝑆𝑉

𝑏𝑉

𝑐�� =394.2

27.22

��𝑽 = 𝟏𝟒. 𝟒𝟖 ft

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tan(𝛬𝑉) =𝐶

𝑏𝑣

C = bV ∗ tan(ΛV)

𝐶 = 27.22 ∗ tan(35°)

𝐶 = 19.06 ft

𝐶 = 𝐶𝑣,𝑟𝑜𝑜𝑡 − 𝐶𝑣,𝑡𝑖𝑝

𝐶 = 𝐶𝑣,𝑟𝑜𝑜𝑡 − 𝜆 ∗ 𝐶𝑣,𝑟𝑜𝑜𝑡

𝐶 = 𝐶𝑣,𝑟𝑜𝑜𝑡 ∗ (1 − 𝜆)

𝐶𝑣,𝑟𝑜𝑜𝑡 =2 ∗ 𝑆𝑉

𝑏𝑉 ∗ (1 + 𝜆)

𝐶 =2 ∗ 𝑆𝑉

𝑏𝑉 ∗ (1 + 𝜆)∗ (1 − 𝜆)

𝜆 =2 ∗ 𝑆𝑉 − 𝐶 ∗ 𝑏𝑉

2 ∗ 𝑆𝑉 + 𝐶 ∗ 𝑏𝑉

𝜆 =2 ∗ 394.2 − 19.06 ∗ 27.22

2 ∗ 394.2 + 19.06 ∗ 27.22

𝝀 = 𝟎. 𝟐𝟎𝟔

𝐶𝑣,𝑟𝑜𝑜𝑡 =𝐶

(1 − 𝜆)

𝐶𝑣,𝑟𝑜𝑜𝑡 =19.06

(1 − 0.206)

𝑪𝒗,𝒓𝒐𝒐𝒕 = 𝟐𝟒. 𝟎𝟎 ft

𝐶𝑣,𝑡𝑖𝑝 = 𝜆 ∗ 𝐶𝑣,𝑟𝑜𝑜𝑡

𝐶𝑣,𝑡𝑖𝑝 = 0.206 ∗ 24

𝑪𝒗,𝒕𝒊𝒑 = 𝟒. 𝟗𝟓 ft

The values of the Boeing 737-100’s vertical stabilizer were used to calculate the dimensions of

the A320neo’s stabilizer. These are reasonable values that align with the size of our aircraft.

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Horizontal Stabilizer

The horizontal stabilizer was defined using comparisons from the Fokker 100 jet transport.

The following equations are accepted ratios used to estimate the geometry of an aircraft’s

stabilizer.

��𝐻 =𝑙𝑆ℎ

𝐶𝑆

��𝑯 = 𝟏. 𝟎𝟕

𝑙

𝐿= 0.5

𝐿 = 123.25 → 𝒍 = 𝟔𝟏. 𝟔𝟐𝟓 ft

𝑺𝑯 =1.07∗13.8∗1460

61.625= 𝟑𝟒𝟗. 𝟖𝟑 ft2

𝐶𝐻 = 0.6 ∗ 𝐶��𝑖𝑛𝑔 𝐶𝐻 = 0.6 ∗ 13.8

𝑪𝑯 = 𝟖. 𝟐𝟖 ft

𝑏𝐻 = 𝑆𝐻

𝐶��

𝑏𝐻 = 349.83

8.28

𝒃𝑯 = 𝟒𝟐. 𝟐𝟓 ft

𝐶𝑟𝑜𝑜𝑡 = 2𝑆𝐻

𝑏𝐻(1 + 𝜆)

��𝒓𝒐𝒐𝒕 =2∗349.83

42.25(1+0.27) = 13.04 ft

𝐶𝑡𝑖𝑝 = 𝜆𝐶��𝑜𝑜𝑡

��𝒕𝒊𝒑 = 0.27 ∗ 13.039 = 𝟑. 𝟓𝟐 ft

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Stabilizer Dimensions

Characteristic Variable Vertical

Stabilizer Value

Horizontal

Stabilizer Value

Area 𝑆𝑉, 𝑆𝐻 394.2 ft2 349.8 ft2

Mean Aerodynamic

Chord 𝐶𝑉 , 𝐶𝐻 14.48 ft 8.28 ft

Span 𝑏𝑉 , 𝑏𝐻 27.22 ft 42.25 ft

Taper Ratio 𝜆𝑉 , 𝜆𝐻 0.206 0.27

Root Chord 𝐶𝑉,𝑟 , 𝐶𝐻,𝑟 24.00 ft 13.04 ft

Tip Chord 𝐶𝑉,𝑡, 𝐶𝐻,𝑡 4.95 ft 3.52 ft

VI. SolidWorks Model

In order to better understand the information we are collecting for this airplane, a SolidWorks

CAD model was created to run several aerodynamic simulations on the plane. This will allow us

to have experimental data to compare with our mathematical numbers. To get the best results

possible we began with a detailed reference CAD model, then to maximize efficiency the design

was modified to reduce features that we would not need specific data from. The first major

change was removing the engines, which were not a part of our aerodynamic study.

Official CAD Model (used for reference)

18

Seen above is the actual Solidworks model of the Airbus A320neo. The two glaring differences

are the differing shape of the wings and the flat fuselage section at the wing interface. The wings

on this model have less of a taper along with a reduced sweep angle as they approach the fuselage.

This model also has winglets and vortex generators which are not accounted for in our model. The

stabilizer on the actual model has a variable sweep while the team model uses straight sweeps.

Team Model (used for flow simulation)

When comparing the fuselages between both models, there are several clear differences. In the

reference model, the fuselage has two flat side pieces that help with the connection of the wings.

The fuselage also has landing gear, doors and windows built into the model. The plane also models

the PW1100 jets and features various avionics equipment. For the team model, the fuselage is a

straight round tube for most of the length.

VII. CFD Simulation

CFD Simulations were used to gather another set of data from to compare to hand calculations

focused on the drag and lift forces. The flow simulations were conducted using SolidWorks.

Simulations were done on a modified scale model of the aircraft to reduce the computation power

required to obtain clean results.

Hand Calculations

Wing Method

The calculations for the parasitic and skin friction coefficient are shown below.

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𝑅𝑒 = 𝜌∞𝑣∞𝑚𝑎𝑐

𝑢∞

𝑅𝑒 = (0.000738)(765)(13.80)

(2.995 ∙ 10−7)

𝑅𝑒 = 2.601 ∙ 107

𝐶𝑓 =

0.455

𝑙𝑜𝑔(𝑅𝑒)2.58−

1700

𝑅𝑒

𝐶𝑓 =

0.455

𝑙𝑜𝑔(2.601 ∙ 107)2.58−

1700

2.601 ∙ 107

𝐶𝑓 = 0.002523

𝑚𝑎𝑐 = 13.8

𝑆𝑤𝑒𝑡 = 2(1 +0.2𝑡

𝑐)𝑆𝑒𝑥𝑝𝑜𝑠𝑒𝑑

𝑆𝑤𝑒𝑡 = 2 (1 +0.2(1.66)

13.8) 1170

𝑆𝑤𝑒𝑡 = 2396.29

𝐶𝐷0,𝑊𝑖𝑛𝑔 = 𝑘𝐶𝑓

𝑆𝑤𝑒𝑡

𝑆𝑟𝑒𝑓

𝐶𝐷0,𝑊𝑖𝑛𝑔 = (1.22)(0.002523)(2396.29)

205.9

𝐶𝐷0,𝑊𝑖𝑛𝑔 = 0.0358

Stabilizer Method

𝑆𝑤𝑒𝑡 = 2(1 +0.2𝑡

𝑐)(𝑆𝑒𝑥𝑝𝑜𝑠𝑒𝑑)

𝑆𝑤𝑒𝑡 = 2(1 +0.2(1.66)

13.8)(263)

𝑆𝑤𝑒𝑡 = 537.6

𝐶𝐷0,𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑧𝑒𝑟 = 𝑘𝐶𝑓

𝑆𝑤𝑒𝑡

𝑆𝑟𝑒𝑓

𝐶𝐷0,𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑧𝑒𝑟 = (1.22)(0.002523)(537.6)

31.9

20

𝐶𝐷0,𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑧𝑒𝑟 = 0.052

Fuse Method

𝑆𝑊𝑒𝑡,𝑁𝑜𝑠𝑒 = 0.75 ⋅ πD𝐿𝑁𝑜𝑠𝑒

𝑆𝑊𝑒𝑡,𝑁𝑜𝑠𝑒 = 0.75 ⋅ π(12.96)(9)

𝑆𝑊𝑒𝑡,𝑁𝑜𝑠𝑒 = 274.826

𝑆𝑊𝑒𝑡,𝐵𝑜𝑑𝑦 = πD𝐿𝐵𝑜𝑑𝑦

𝑆𝑊𝑒𝑡,𝐵𝑜𝑑𝑦 = π(12.96)(99.8)

𝑆𝑊𝑒𝑡,𝐵𝑜𝑑𝑦 = 4063.24

𝑆𝑊𝑒𝑡,𝑇𝑎𝑖𝑙 = πD𝐿𝑇𝑎𝑖𝑙

𝑆𝑊𝑒𝑡,𝑇𝑎𝑖𝑙 = π(12.96)(14.5)

𝑆𝑊𝑒𝑡,𝑇𝑎𝑖𝑙 = 590.3

𝑆𝑤𝑒𝑡 = 𝑆𝑊𝑒𝑡,𝑁𝑜𝑠𝑒 + 𝑆𝑊𝑒𝑡,𝐵𝑜𝑑𝑦 + 𝑆𝑊𝑒𝑡,𝑇𝑎𝑖𝑙

𝑆𝑤𝑒𝑡 = 274.83 + 4063.24 + 590.3

𝑆𝑤𝑒𝑡 = 4928.37

𝐶𝐷0,𝐹𝑢𝑠𝑒 = 𝑘𝐶𝑓

𝑆𝑤𝑒𝑡

𝑆𝑟𝑒𝑓

𝐶𝐷0,𝐹𝑢𝑠𝑒 = (1.15)(0.002523)(4928.37)

131.92

𝐶𝐷0,𝐹𝑢𝑠𝑒 = 0.108

𝐶𝐷0 = 𝐶𝐷0,𝑊𝑖𝑛𝑔 + 𝐶𝐷0,𝐹𝑢𝑠𝑒𝑙𝑎𝑔𝑒 + 𝐶𝐷0,𝑆𝑡𝑎𝑏𝑖𝑙𝑖𝑧𝑒𝑟

𝐶𝐷0 = 0.0358 + 0.108 + 0.052

𝐶𝐷0 = 0.1958

Obtaining Lift Coefficient:

From the 𝐶𝐿 vs 𝛼 chart, 𝑎0 = 0.1049 for an infinite wing (per degrees). Calculating 𝑎 for a finite

wing:

𝑎 =𝑎0

1 +57.3 ∗ 𝑎0

𝜋(𝐴𝑅)𝑒

21

𝑎 =0.1049

1 +57.3 ∗ 0.1049𝜋(9.45)(0.85)

𝑎 = 0.0847

From the 𝐶𝐿 vs 𝛼 chart, 𝛼𝐿=0 = −3,

𝐶𝐿 = 𝑎 ∗ (𝛼 − 𝛼𝐿=0)

𝐶𝐿 = 0.0847 ∗ (0 − (−3))

𝐶𝐿 = 0.254

Calculating Induced Drag coefficient 𝐶𝐷𝑖 ,

𝐶𝐷𝑖 =𝐶𝐿

2

𝜋(𝐴𝑅)𝑒

𝐶𝐷𝑖 =𝐶𝐿

2

𝜋(9.45)(0.85)

𝐶𝐷𝑖 = 0.00256

Calculating total drag coefficient 𝐶𝐷 ,

𝐶𝐷 = 𝐶𝐷𝑖 + 𝐶𝐷0

𝐶𝐷 = 0.00256 + 0.1958

𝐶𝐷 = 0.1984

Calculating lift to drag ratio 𝐶𝐿

𝐶𝐷 ,

𝐶𝐿

𝐶𝐷 =

0.254

0.1985

𝐶𝐿

𝐶𝐷 = 1.28

After obtaining the coefficient of drag for the entire plane, the drag force can be calculated.

𝐷 =1

2𝜌∞𝑣∞

2𝐶𝐷 𝑆𝑤𝑒𝑡

𝐷 =1

2(0.000738)(765)2(0.009353)(7862.3)

𝐷 = 15880 lb

Calculating total lift,

𝐿 = 𝐷 ∗𝐶𝐿

𝐶𝐷 = 15880 ∗ 1.28

𝐿 = 20326 lb

22

These calculations must be repeated for each of the desired angles of attack.

Angle of

Attack (𝛼, deg)

Lift

Coefficient

Drag

Coefficient

Lift Drag

Ratio Lift (lb) Drag (lb)

0 0.254 0.198 1.28 20326 15880

5 0.678 0.214 3.17 54407 17163

10 1.101 0.244 4.51 88258 19569

CFD Simulation

The plane was simulated in SolidWorks Flow Simulation with a scaled model (b = 1ft). The

first image shows the size of the computational domain which only contains half of the plane. The

squares in dark blow show that basic mesh. Lighter colors show the refinement level (up to 7). The

bottom image shows the basic mesh and refinement on the wings and vertical stabilizer.

23

These images show the detailed refinement at the planes lift and control surfaces. The mesh has

490,000 cells and 241,000 fluid cells along the solid border, indicating a detailed mesh at in the

boundary layer. For each angle of attack, the boundary conditions are changed to change the

direction of flow.

24

0˚ Angle of Attack

25

5˚ Angle of Attack

26

10˚ Angle of Attack

27

CFD Results

0˚ Angle of Attack

Variable Value

𝐶𝐿 0.00776

𝐶𝐷 0.00167

𝐶𝐿 𝐶𝐷⁄ 4.66

5˚ Angle of Attack

Variable Value

𝐶𝐿 0.01341

𝐶𝐷 0.00283

𝐶𝐿 𝐶𝐷⁄ 4.73

10˚ Angle of Attack

Variable Value

𝐶𝐿 0.01582

𝐶𝐷 0.00449

𝐶𝐿 𝐶𝐷⁄ 3.52

The 𝐶𝐿 𝐶𝐷⁄ ratio drops off at 10˚ because there is a separation of flow. As expected, the drag

coefficient increases for each trial. At 10˚, the lift coefficient drops off slightly indicated the flow

separation.

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VIII. Conclusion

This report of Analysis of Airbus A320 concludes a discrepancy of approximately 33% for the

5 degree angle of attack and 22% for the 10 degree angle of attack between the calculated values

and the simulated values of the lift to drag coefficients ratio. The leading causes of this discrepancy

are the differences in the model used for the simulations and the real aircraft, and the inability of

basic equations to proper calculate at the level of a flow simulation.

There are a couple of differences between the actual SolidWorks model of the Airbus A320neo

and the model made through calculated dimensions. The two glaring differences are the differing

shape of the wings and the flat fuselage section at the wing interface. The wings on this model

have less of a taper along with a reduced sweep angle as they approach the fuselage. This model

also has winglets and vortex generators which are not accounted for in our model. The stabilizer

on the actual model has a variable sweep while the team model uses straight sweeps. When

comparing the fuselages between both models, there are several clear differences. In the reference

model, the fuselage has two flat side pieces that help with the connection of the wings. The fuselage

also has landing gear, doors and windows built into the model. The plane also models the PW1100

jets and features various avionics equipment. For the team model, the fuselage is a straight round

tube for most of the length.

29

IX. References

[1] “A320. AIRCRAFT CHARACTERISTICS. AIRPORT AND MAINTENANCE

PLANNING.” Airbus. Accessed 8 June 2020. Website.

https://www.airbus.com/content/dam/corporate-

topics/publications/backgrounders/techdata/aircraft_characteristics/Airbus-

Commercial-Aircraft-AC-A320.pdf

[2] “A320: by Airbus”. Aircraft Performance Database. Accessed 12 May 2020. Website.

https://contentzone.eurocontrol.int/aircraftperformance/details.aspx?ICAO=A320

[3] “A320neo: Unbeatable Fuel Efficiency”. AIRBUS. Accessed 12 May 2020. Website.

https://www.airbus.com/aircraft/passenger-aircraft/a320-family/a320neo.html

[4] Avinash and Jahnavi. “Aircraft Design and Weight Estimation Nomenclature”. Global

Journal of Researches in Engineering: B Automotive Engineering. Volume 14, Issue

4. Published 2014. Website. https://globaljournals.org/GJRE_Volume14/4-Aircraft-

Design-and-Weight.pdf

[5] Hensey and Magdalina. “A320 NEO vs. CEO comparison study”. FPG Amentum.

Published 19 July 2018. Technical Report. https://www.fpg-amentum.aero/wp-

content/uploads/2018/07/180719-FPG-Amentum-research-A320-NEO-vs-CEO-

comparison-study.pdf

[6] “NACA 63-412 AIRFOIL (n63412-il)”. AirfoilTools.com. Accessed 9 June 2020.

Website. http://airfoiltools.com/airfoil/details?airfoil=n63412-il#polars

[7] Roy, Shaylesh. “AIRBUS A320neo”. GRABCAD.com. Published 24 February 2020. 3D

CAD Model. https://grabcad.com/library/airbus-a320neo-1