Low Speed Aerodynamics
Low Speed Aerodynamics AE 2020
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Date Created: 11/02/15
AE2020 Low Speed Aerodynamics Fall 2007 Lecture 1 Introduction Narayanan Komerath Professor Daniel Guggenheim School of Aerospace Engineering Georgia Institute of Technology Atlanta Notes Email Komerathgatechedu I check email a lot more frequently than I check my of ce voicemail 4048943017 Of ce Hours TBA Tentatively 89 TueThu Text Either Fundamental of Aerodynamicsquot by Anderson or Aerodynamics for Engineersquot by Bertin and Smith Attendance is expected as much as in a workplace 439 r A flumi A rst tL i l I 39 r AERODYNAMICS 39Lifebaaf Lands 5339er Edwavds CaMovma Vest h dm d Asecdndsmm Howevevhe m esanhom L some dam A We ve gomg wmpmve We deswgn he added emevgency mudman hq nasa ggvmnovahonmnovatmLmAmagesphotgsX38 pg FORCE BALANCE lN FLIGHT The 39 a J L horizontal quot 39 subsonic s eeds The rest ofthe aircrait just hangs from these quotlilting surfacesquot Of course the wings and tails themselves have weig t On most aircra the wings contain most ofthe fuel the weight of the aircra at Irr quot quotawsof quot 39 quot quot 39 quot ofanaircra andthusto 44 A tL ml Li IA iit JJJt 1 Wiiigvv V DlStrlbutlon r Straight And Level Steady Flight In Straight and Level Steady Flight where all the accelerations are zero Li Weight and Thrust Drag L W T D 2 lt THRUST i VWEIGHT in at Basic Concepts and Results in Aerodynamics Freestream Vector Velocity ofthe uid far ahead ofthe object in the ow undisturbed by the presence of the objec rnp Lift Drag A U aircraft U wind Aerodynamic Lift Force perpendicular to the freestream exerted by the flow on an object Drag Force along the freestream acting on the aircraft Drag dynamic pressurepanform areadrag coef cient L TM Li Flight Dir 30 deg G N TliiLisi l D W For straight and level ight LW and TD Weight Flight Dii Horizontal Planform Area S S is the area enclosed by a fullscale drawing of the outline of the wings when viewed from directly above or below Dynamic Pressure in lowspeed flow 1 2 007 U00 4 23 q 1 MW2 psf Lift dynamic pressure planform area lift coef cient The lift coefficient depends on how the 1 lift was generated CL is usually around 01 to 14 L E pUOOZSCL LqooSCL G75g Li Generation Generationur quot 39fromM A v 39 quot motion 1 r When owing air is induced to turn there is a rate of change ofmomentum along the direction perpendicular to the original direction The net force which is required to cause this change causes an equal and opposite reaction on the object which turns the TL39 quot 39 L force quot g quot 39339 Most simply quot quot 439 passive methods to turn the ow Rde of change of momentum Force exerted on uid gt gt gt gt p Reaction exerted by uid Li D splay Slreamhne httpwwwwindsokansascomlAZZFigSjpg How to turn the flow and generate lift Camber A A Examples of Streamline Patterns mm Wva am asn auluvuundschuuvweamhnEs mu Suuvce mm Nuneme newnesmmouwmsmoanen m1 x ms Dovmwash 25 quot39 POTENTIAL FLOW i Band on Garrett Flg 56 Evolution ofthe Ain oil Shape Flat Plate Cambered Plate to delay stall KM Rounded leading edge gives largerrange of angle of attack but blunt trailing edge causes stall and is ineffective in ow turning Flat Plate Stall High drag too Rounded leading edge to delay stall sharp trailing edge to turn the attached ow 6y httpll Glenn Lift from Flow Turning Research Center it is a force Force mass X acceleration m Force mass X change in velocity with time FmV Vn l 1 Velocity has both magnitude speed and direction Changing either the speed or direction of a ow generates a force Liit is a force generated by tuming a moving uid This looks cool but hoW much quotInquot gets turned per unit time How much does it turn These are not simple to predict by looking at the shape a VortexInduced Lift VortexInduced Lift and Delta Wings HrghrAsport Rana ng WE Wm Airfoil British 39aerofoil39 Airfoil means shape ofa section ofa wingquot It is a twodimensional concept Airfoils cannot y wings y Airfoil 439 IA It 4 a Airfoil lift coef cient 639 varies with angle of attack a Ifthe airfoil is cambered the li coef cient is positive even at zero angle of attack and reaches zero only at some negative value of angle ofatack this is called the quotzeroli angle of attackquot an As the camber is increased 0 becomes more negative dc Thus airfoil lift coef cIent Is 5 0 dc The I39 39curVe Slope tells us howmuch li increase we can expect to get 039 from increasing angle of attack from a given angle of attack dc I 5 211 where D is in radians d a G75g Ain oils versus Wings 1539 Lg perm Spain gm c 01 D39 qug permit Span gen r ca 0 Ain oil Chord which is the line joining leading edge and trailing edge CL Stall Note Almost straight line LiftCurve Sloge of an airfoil an owl av Thinairfoil theoryquot in lowspeed flows where Mach number is close to 0 proves that dc 1 3 Zn doc in lo W speed o W5 ExerciseLift coefficient The angle of attack of an airfoil is 12 degrees The lift curve slope is 58 per radian Zerolift angle of attack is 2 degrees Find the lift coefficient If the air density is 110 of sealevel standard and the temperature is 20 deg C higher than the standard sealevel flight speed is 100 ms and wing planform area is 30m find the lift Pressure Coefficient The pressure coefficient is a way to express the pressure with respect to some reference pressure as a quotdimensionlessquot quantity p19 P Pco 0p 5 2 05erm Goo U 2 1 w U Cp 0 indicates the undisturbed freestream value of static pressure Cp 1 indicates a stagnation point Cp lt 0 indicates a suction region Chordwise ressure distribution over an airfoil in lows eed flow P lu p2u Psu l F 2I P3 h 3 Flat Plate Streamlines and Pressure Distribu VIEW Edge 3 O 15 degrees Dlsplay Qtreamllnes Bum Free Stream 103325 tut Ea 8373 99539 E5 93007 ET 85235 5 025 on RES n Fig 5 5 Vehord ExerciseP ressure Coefficient What is the pressure coefficient at the stagnation point of an airfoil section What is the pressure coefficient on a flat surface aligned with the freestream Cp at the suction peak of an airfoil is 12 What is the pressure there as a percentage of the freestream static pressure What is the velocity at this point as a percentage of the freestream velocity Ideally nothing happens to the uid when an object goes through except that it gets moved aside or maybe turned a bit but then recovers Drag In practice some effect always remains Some ow gets pulled along the direction ofthe object until its kinetic energy is dissipated as heat rthere is a region of swirling ow le behind whose energy is also dissipated as heat The force resulting from these things is drag It acts along the freestream direction Th d 39 39 b e rag Is given y D qOOSCD Li to Drug Ratio LD 37 We want our airplanes to have as high LID as possible Most aircraft are designed to mlnllcm The profile drag of an airfoil of chord 1unit is about the same as that of a circular cylinder whose diameter is only 0005 units U Airfoil of chord c U 3 Cylinderof diameter 00050 Streamlining At high Reynolds numbers the drag of an airfoil is chord c is only as much as that of a cylinder of diameter 00050 From Covert Eugene quotThe Legacy ofthe Wright Brothers and Its Futurequot 1997 Wright Broth ers Lecture a But we want our space reentry capsules and parachutes to have very low LD hMpNanasa gov 330199 Drag Coefficient The drag is given lD USpUmZSCE The drag coefficient in lowspeed flow is composed of 3 parts CD ODO CDfrr39c on ODr wherICDg is the parasite drag which is independent of lift It is usually due to the losses of stagnation pressure which occur when part of the flow separates somewhere along the wing or body surface In high speed flight the effect of shocks and wave drag must be added to this and becomes the dominant source of drag Example cm of a small airliner is 0018 Wing aspect ratio is 6 Assume spanwise Efficiency is 10 Lift coefficient is 05 Find the total drag coefficient If the density is 1 kgm3 and speed is 200ms find the drag I we D I 2 is called Equivalent Flat Plate Areaquot qoo Note Unlike lift quot39 quotr 39 39 withthe a r nt is calculated the drag coe produces drag So when the drag of each compone based on the reference areaquot forthat component For instance for a fuselage the reference area is 39 S everything f cient of that is the crosssection area Thus when everything is summed up you use the concept of Equivalent Flat Plate Areaquot rather than s GE CD to make sure we don t use different reference area Example Subscript 0quot indicates zeroliltquot value Component CD0 Area DUIq sqft selage 003 25 area 64 sqft 2 engine nacelles Wings 0003 Planform area 120 sqft 2 Tails 0004 tail planform area 15 sqft 2 Landing Gear 12 strut diameter length 212 5 2 Total Dolq sqft ofthe airplane CD0 based on planform area Let s say that the airplane is ying at minimum drag Liltinduced drag pro le drag Let s say that the aircra weight is 5000 lbs What is the LID LIFTINDUCED DRAG and ASPECT RATIO At the ends of the wings the pressure difference between the upper and lower sides is lost as the flow ros up into a vortex Inboard vortex sheet Tipvo rtices 2 AR I S where b is the wing span and S is the I wing planform area Span b The Aspect Ratio of a wing is defined as eff Effects of Finite Aspect Ratio 1 The overall li is reduced relative to the airfoil li value predicted for a section of an in nite wing 2 The li vector is tilted back so that an quotinduced dragquot is created Drag D Lift L Normal Force F Vinf Vind Veff Both ofthese u uany 39 effects are inuea ing in Aspect Ratio of the wing G75g Points to Note on Finite Wings 1 Since induced drag is directly related to the li it can be calculated by the same mathematical formulation used to calculate li 2 Does not require consideration ofviscosity 3 No induced drag on 2D airfoils under steady conditions 4 At nite Aspect ratios C 2 L C 2 Dl 7rARe 5 Ideal elliptic li distribution implies minimum induced drag ie spsnwise ef ciency 5 1 7 Note lnthe2Dlimit quotquot quot 39 quotquot 4 4 39 39 39 uw pro le dagdu t 39 quot39 4 a I 39 I drag 3 Effect of Aspect Ratio on Wing Lift Curve Slope dim 6390 d 1 aAR 11 1T doc Transport aircraft LD 16 N 20 Fighters LDcruise 10 16 Supersonic transport LD 11 Hypersonic aircraft LD 1 lift is a minor problem here Example An L a a 3 r r 1 1 ML n aspect ratio of 7667 and wing span of 6096m We39ll assume that its spanwise ef ciency factor will be 099 Let39s assume that the pro le drag coe l cient is given Y CD0 0015 Thus formaximum LilttoDrag ratio minimum drag and the li is always equal to the weight for straight and level ight on cm 0015 The corresponding CL is calculated as 0598 and the dynamic pressure is 101123me At 11000 meters in the Standard Atmosphere density is 036kglm3 so that the ight speed is 24064 mIs Note In practice the CD0 might change with ight Mach number for highspeed ight This is not taken into account in the above Li 11 n A J Stereolithog raphy models of the NASA X38 Georgia Institute of Technology School ofAerospace Engmq AERODYNAMICS SUMMARY i is force perpendicular to the ow direction due to pressure differences across surfaces 3 ways of generating Ii a angle of attack b camber c vortexinduced Ii An in nite 2dimensional wing is entirely described by its airfoil section Finite wings have less Ii than corresponding spanlengths of an in nite wing at the same angle of attack and also have Ii induced drag The total drag is composed of pro le drag which does not vary with Ii and induced drag which rises as the square ofthe Ii coef cient To y an airplane of a given weight straight and level the condition for minimum drag maximum lifttodrag ratio is that the pro le drag coef cient is the same as the induced drag coe icient The slope ofthe liltcurve above is indeed nearly constant over the quotsmallangleofattackquot region from about 5 to 10 degrees for most airfoils In other words a plot of lilt coef cient versus angle of attack is straight until one reaches near the stalling angle of attack The maximum slope ofthis line theoretical limit for a thin airfoil at small angle of attack meaning below about10 degrees is 21m lfDL is in radians 011 2 f do 7 39 This is a neat result Now you can amaze your iends by calculating the approximate lilt coef cient of an airfoil at agiven angle of attack simply by multiplying the degrees by 7 and dividing by 180 to convert to degrees and then multiplying this by 2times1r which is 628 Hint Trsquared is 99696 Can you remember that approximately Also nd 2nsquaredl180 Can you remember that approximately 21 is the quotideal li curve slopequot predicted by quotthin airfoil theoryquot as we will see later in this course People can design airfoils that come fairly close to this value So 01 7 39i39 c 11b 0L0 is the value OfDL which gives cl 0 with on in radians G is the skin friction drag which is due to Cgric on viscosity is the Induced Drag In lowspeed flight this is the largest cause of drag because you have to have lift to fly and this drag is caused by lift Here e is called the quotspanwise efficiency factorquot It is C L the answer to the question How does this wing rate D 11 AR jg compared to the ideal wing for this aspect ratio Its value is usually close to 1 perhaps as high as 099 CD Note tha CD ac Ci 2 somatcncca AlsoCD U as AR gtUrl To minimize induced drag one should design wings with the largest possible aspect ratio but also provide enough surface area so that you need only a small angle of attack to provide the necessary lift even at low speed me andesnrgukandes h 39J information m escondor punnZou jpg Speed for Minimum Drag Total drag is composed of a part which depends on lift and one that does not 2 D DDDi CD0CDU5pr 3 6 2 D CD0 MARE 053000 5 Let us consider what it takes to keep L W H WLquCL CL r1055quot W 2 1 S D gamma 1iAReZD dB E 2 1 i a SOLD lt5 MARRLQ W Le Cm 09 Or Minimum Total Drag twice zeroilift drag So This is a remarkable result It means that AIRCRAFT UNLIKE OTHER FORMS OF TRANSPORTATION HAVE A DEFINI TE SPEED FOR MINIMUM DRAG 39 DzDiDO o I Drag o Di s 60 9min a u i I39 l I Speed jr minimum mag Flight Speed U To fly an airplane of a given weight straight and level the condition for minimum drag maximum Iifttodrag ratio is that the pro le drag coef cient is the same as the induced drag coef cient a AE2020 Low Speed Aerodynamics Fall 2007 Lecture 2 Fluid Mechanics Introduction Copyghf2007 N Kaineam Other rights may be specNed mm indvdua items Ail Ng ls reserved Types of Fluid Motion 1 Fluids like most other forms of matter are made up oftiny particles molecules which are separated by large spaces Copyig f2007 N Kaineam Other rights may be specnee win indvidual items All righls reserved Avogadro39s is 60221415 x 1023 gives of atoms or molecules per grammole of the gas nun many L I quotI r lgrammole of air is roughly 2897 grams Molecularweight Mis 2897 gramsmole or kglkgmole Density of air massvolume can be found from the Perfect Gas Law Lets say we have air at sea level standard conditions Pressure is 1013 x 105 Nlm2 and temperature is 288K P p7 A 71013251 2897 piPM RT 8313 288 RT Density is 1225 kglm3 So how many grams in a cubic centimeter 1225108 73 How many molecules 12 170 60221415 gt 023 255 1019 So we can treat uid as a quotcontinuumquot a medium which is pretty uniform in properties at the smallest scales of interest to us Copyghf2007 N Kaineam Other Hg fs may be specNed mm indwow items All Ng ls reserved d dxfdyfdzi Zu3viwi lfds is an elemental vector along the streamline 1 J k ds xl dx dy dz u v w 31Wdyivd2judzi wdxkvdxiudy 0 Equation to a streamline is wdyvdz K dy v udzwdx 622 six 4 Copyght2007 N Kaineam Other Hg fs may be specNed mm mdva ual ltems All ngms reserved Four Basic Types of Fluid Motion As a quotpacketquot of uid moves along a combination of4things can happen to it translation dilatation rotation and shear I J L J r AL Thus any 1Translation motion ofthe center ofmass This is characterized by the velocity 5 u I v j w l p 7 P RT 2 Dilatation volume change 3 Rotation About one two or 3 axes 4 Shear rain Copyghf2007 N Karrieam Other fg fs may be specNed mm mdva ua Uems All ngms reserved 2 Dilatation volume change a u 0v 0w Vou 777 0x 0y 02 Also called Divergence ofthe velocity vectorquot oepw r2oo7v Kaineam Omerng tsmay be specNed mm indwow items All Ng ls reserved 3 Rotation About one two or3 axes 1 dw dv du dw d1 du 007 iiil fiijJr iii 2 dy dz a2 dx dx dy vorzicz39fy 253 V x z Note that rotation is a vector We usually use a quantity called quotvorticityquot which is twice the rotation vector to describe the amount of rotation in ows Copyght2007 N Kaineam Other Hg fs may be specNed mm indwow items All Ng ls reserved ll 4 Shear Strain Strain is de ned as a change in length per unit length Rate of strain is thus given by change in velocity per unit len th Shear exists when there is a gradient ofvelocity along the direction perpendicular to streamline g 9 E aw av iau 0w 9 a 2 if i x 6y y 6y 62 62 6x The quantity quotstrain ratequot has nine components They are exx eyy ezz These are quotnormal strain ratequot components exy eyz ezx These are quotshear strain ratequot components eyx ezy exz These are equal to the corresponding quotshear strain ratequot components above Copyght2007 N Kaineam Other Hg fs may be specNed mm indwow items All Ng ls reserved rm a W Circulation F L7 0d Defined as integrated around a closed contour C The negative sign is included such that positive circulation on a body corresponds to positive lift and the integral is evaluated counterclockwiseFrom the preceding discussion we see that will be zero unless there is some vorticity contained within the contour r is an extremely useful quantity it helps us calculate lift vortex strength etc Important Points The circulation around a closed contour with net rotation andor shear will be nonzero It is however always possible to have a combination of rotation andor shear that gives a zero circulation Note In the above the de nition is based on a line gt gt gt gt gt gt u Integral around a contour which is In a given plane y a a g 19 gt This works when we know the plane of the rotation and gt a gt Z a that the flow is entireyZdimensional To deal with gt s 11 39 399 a iii L gt general problems we can considerFto be a vector 9 aligned along the aXis of rotation 4 gt gt Copyright 2007 N K omerath Other rights may be speci ed with individual items All rights resened