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Final Presentation

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In this ppt a mathematical and numerical model of airflow around the airfoils in different flow regimes using G-msh , MATLAB and ANSYS as a tool.
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FLOW AROUND AIRFOILS IN DIFFERENT FLOW REGIMES Team Members: Chandrashekar Vishwanadha Jibi Ninan Varughese Krishna P Nitya Kalva
Transcript
Page 1: Final Presentation

FLOW AROUND AIRFOILS IN DIFFERENT FLOW

REGIMESTeam Members:

Chandrashekar VishwanadhaJibi Ninan Varughese

Krishna PNitya Kalva

Page 2: Final Presentation

ContentsIntroductionProcedureResults: Comparison of different casesConclusionReferences

Page 3: Final Presentation

Objective

To analyze the flow around different airfoils with varying angle of attacks and in different flow regimes

Page 4: Final Presentation

Coordinates of

Airfoil

MATLAB for Gmsh

script generatio

n

Gmsh code

Mesh generation using Gmsh

CFD Analysis

in ANSYS Fluent

Procedure

Page 5: Final Presentation

Meshed Airfoil GeometryAsymmetric Air-foil Symmetric Air-foil

• NACA 0012 airfoil• Max thickness 12% at 30% chord• Max camber 0% at 0% chord

• NACA 6409 airfoil• Max thickness 9% at 29.3% chord• Max camber 6% at 39.6% chord

Page 6: Final Presentation

CFD AnalysisSetup

Solver Boundary conditions

• Type: Density-based• Steady State• 2-D planar space• Absolute velocity

formulation

• Implicit• 2nd Order upwind• Solution steering to

dynamically change CFL• Under-relaxation factor:

0.5

Flow conditions• Inviscid flow• Ideal gas• Energy Equation: On• Po = 1 (atm.), To = 300

( K)

Pressure far field

Pressure far field

Pre

ssu

re f

ar

field P

ressu

re fa

r field

wall

Page 7: Final Presentation

CasesCase 1: Symmetrical Wedge

Mach no = 1.5Mach no = 3

Case 2: Varying Mach numbers @ α = 0o

Symmetrical airfoil (M = 0.2, 1, 3, 7) Asymmetrical airfoil (M = 0.2, 1, 3, 7)

Case 3: Varying Angle of attack @ M =3Symmetrical airfoil (α = 5o, 10o, 15o)Asymmetrical airfoil (α = -15o, 5o, 15o)

Page 8: Final Presentation

ResultsCase 1: Symmetrical Wedge

Page 9: Final Presentation

Mach contour – Mach 1.5 • Detached shock as

expected from the theta- beta-M diagram

Mach contour- Mach 3 • Attached shock• M2= 2.23 which is

approximately equal to theoretical value of (2.26)

• M3=3.64 which is approximately equal to theoretical value 3.65

Wedge: Varying Mach Number

Page 10: Final Presentation

Wedge

Velocity vector plot for Mach 1.4• Sudden change in direction of

flow at bow shock• Gradual change in direction of

flow in Prantdl Meyer expansion

Pressure plot for Mach 1.4

Page 11: Final Presentation

ResultsCase 2: Varying Mach Number

Page 12: Final Presentation

Symmetric Airfoil: Varying Mach NumberLow Subsonic: M =0.2 Transonic: M = 1

Supersonic: M = 3 Hypersonic: M = 7

Mach number contours

Page 13: Final Presentation

Symmetric Airfoil: Varying Mach NumberPressure contours

Low Subsonic: M =0.2 Transonic: M = 1

Supersonic: M = 3 Hypersonic: M = 7

Page 14: Final Presentation

Asymmetric Airfoil: Varying Mach NumberLow Subsonic: M =0.2 Transonic: M = 1

Supersonic: M = 3 Hypersonic: M = 7

Mach number contours

Page 15: Final Presentation

Asymmetric Airfoil: Varying Mach NumberPressure contours

Low Subsonic: M =0.2 Transonic: M = 1

Supersonic: M = 3 Hypersonic: M = 7

Page 16: Final Presentation

ResultsCase 3: Varying Angle of Attack

Page 17: Final Presentation

Symmetric Airfoil: Varying Angle of Attackα = 5o α = 10o

α = 15o

Mach number contours

• Symmetric airfoils produce no lift at zero angle of attack

• velair, top > vel air, bottom

• So, according to Bernoulli's principle (ideal conditions):

Pressuretop < Pressurebottom

• This produces lift!

Slip line

Page 18: Final Presentation

Symmetric Airfoil: Varying Angle of Attackα = 5o α = 10o

α = 15o

Pressure contours

• As angle of attack increases, the differential pressure between top and bottom surfaces increases, thus increasing the lift

Page 19: Final Presentation

Asymmetric Airfoil: Varying Angle of Attackα = 5o α = 15o

α = -15o

Mach number contours

• Asymmetric airfoils produce lift even at zero angle of attack because of difference in surface areas of top and bottom surfaces

• Positive α: velair, top > vel air,

bottom

• Negative α: velair, top < vel air,

bottom

Page 20: Final Presentation

α = 5o α = 15o

α = -15o

Pressure contours

• As angle of attack increases, the differential pressure between top and bottom surfaces increases, thus increasing the lift

• Airfoil is being pushed down during negative angle of attack i.e. lift is negative

Asymmetric Airfoil: Varying Angle of Attack

Page 21: Final Presentation

Lift and drag coefficients

0 5 10 15 20 25 300

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.088

0.09

0.092

0.094

0.096

0.098

0.1Cd,Cl vs AOA

C_lift_symmetric

Angle of attack [degrees]

C_l

ift

C_d

rag

0 1 2 3 4 5 6 7 8

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Cl Cd

Cd, Cl vs Mach Number

Mach number

C_l

ift

C_d

rag

Page 22: Final Presentation

ConclusionThe wedge was perfectly symmetrical and thus

produced no lift (C_lift ~ E-06) with zero angle of attack. It also helped in investigating attached shocks and helped in validating the fluent solver

Flow over symmetrical and asymmetrical airfoils was studied under different sonic conditions and angles of attack

The optimal operating conditions observed in fluent agreed with theoretically specified values

Page 23: Final Presentation

ReferencesModern Compressible Flow: With Historical

Perspective. John D. Anderson, JR, 3rd Edition http://airfoiltools.com/search/indexfluent_13.0_workshop02-airfoil.pdfhttp://en.wikipedia.org/wiki/Airfoil

Thank You!Questions?

Page 24: Final Presentation

AppendixHypersonic Flow


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